RS Aggarwal Solutions for Class 11 Chapter 17 Trigonometric Equations

Exercise 17

Exam Tip: Principal solutions are always the values of x that lie in the interval [0, 2π). Use the standard trigonometric solution formulas and substitute n = 0, 1, 2... to find candidates, then keep only those that fall within this range.

Question 1. Find the principal solutions of each of the following equations:
(i) \( \sin x = \frac{\sqrt{3}}{2} \)
(ii) \( \cos x = \frac{1}{2} \)
(iii) \( \tan x = \sqrt{3} \)
(iv) \( \cot x = \sqrt{3} \)
(v) \( \text{cosec } x = 2 \)
(vi) \( \sec x = \frac{2}{\sqrt{3}} \)
Answer: To find the principal solution, apply the standard solution formulas for each trigonometric function and identify values in the range [0, 2π).

(i) Given: \( \sin x = \frac{\sqrt{3}}{2} \)

Using the formula \( \sin\theta = \sin\alpha \implies \theta = n\pi + (-1)^n \alpha \), we have

\( \sin x = \frac{\sqrt{3}}{2} = \sin\frac{\pi}{3} \implies x = n\pi + (-1)^n \frac{\pi}{3} \)

For \( n = 0 \): \( x = 0 + (-1)^0 \frac{\pi}{3} \implies x = \frac{\pi}{3} \)

For \( n = 1 \): \( x = 1 \times \pi + (-1)^1 \frac{\pi}{3} \implies x = \pi - \frac{\pi}{3} = \frac{2\pi}{3} \)

Principal solutions: \( x = \frac{\pi}{3} \) and \( \frac{2\pi}{3} \)

(ii) Given: \( \cos x = \frac{1}{2} \)

Using the formula \( \cos\theta = \cos\alpha \implies \theta = 2n\pi \pm \alpha \), we have

\( \cos x = \frac{1}{2} = \cos\frac{\pi}{3} \implies x = 2n\pi \pm \frac{\pi}{3} \)

For \( n = 0 \): \( x = 0 \pm \frac{\pi}{3} \implies x = \frac{\pi}{3} \)

For \( n = 1 \): \( x = 2\pi \pm \frac{\pi}{3} \implies x = \frac{5\pi}{3}, \frac{7\pi}{3} \)

\( \frac{7\pi}{3} > 2\pi \), so it is not included in the principal solution.

Principal solutions: \( x = \frac{\pi}{3} \) and \( \frac{5\pi}{3} \)

(iii) Given: \( \tan x = \sqrt{3} \)

We know that \( \tan\theta \times \cot\theta = 1 \)

Using the formula \( \tan\theta = \tan\alpha \implies \theta = n\pi \pm \alpha \), we have

\( \tan x = \sqrt{3} = \tan\frac{\pi}{3} \implies x = n\pi + \alpha \)

For \( n = 0 \): \( x = n\pi + \frac{\pi}{3} \implies x = \frac{\pi}{3} \)

For \( n = 1 \): \( x = \pi + \frac{\pi}{3} \implies x = \frac{4\pi}{3} \)

Principal solutions: \( x = \frac{\pi}{3} \) and \( \frac{4\pi}{3} \)

(iv) Given: \( \cot x = \sqrt{3} \)

We know that \( \tan\theta \times \cot\theta = 1 \)

So \( \cot x = \sqrt{3} \implies \tan x = \frac{1}{\sqrt{3}} \)

Using the formula \( \tan\theta = \tan\alpha \implies \theta = n\pi \pm \alpha \), we have

\( \tan x = \frac{1}{\sqrt{3}} = \tan\frac{\pi}{6} \implies \theta = n\pi + \alpha \)

For \( n = 0 \): \( x = n\pi + \frac{\pi}{6} \implies x = \frac{\pi}{6} \)

For \( n = 1 \): \( x = \pi + \frac{\pi}{6} \implies x = \frac{7\pi}{6} \)

Principal solutions: \( x = \frac{\pi}{6} \) and \( \frac{7\pi}{6} \)

(v) Given: \( \text{cosec } x = 2 \)

We know that \( \text{cosec}\theta \times \sin\theta = 1 \)

So \( \sin x = \frac{1}{2} \)

Using the formula \( \sin\theta = \sin\alpha \implies \theta = n\pi + (-1)^n\alpha \), we have

\( \sin x = \frac{1}{2} = \sin\frac{\pi}{6} \implies \theta = n\pi + \frac{\pi}{6}(-1)^n \)

For \( n = 0 \): \( \theta = 0 \times \pi + \frac{\pi}{6}(-1)^0 \implies \theta = \frac{\pi}{6} \)

For \( n = 1 \): \( \theta = 1 \times \pi + \frac{\pi}{6}(-1)^1 \implies \theta = \pi - \frac{\pi}{6} = \frac{5\pi}{6} \)

Principal solutions: \( x = \frac{\pi}{6} \) and \( \frac{5\pi}{6} \)

(vi) Given: \( \sec x = \frac{2}{\sqrt{3}} \)

We know that \( \sec\theta \times \cos\theta = 1 \)

So \( \cos x = \frac{\sqrt{3}}{2} \)

Using the formula \( \cos\theta = \cos\alpha \implies \theta = 2n\pi \pm \alpha \), we have

\( \cos x = \frac{\sqrt{3}}{2} = \cos\frac{\pi}{6} \implies x = 2n\pi \pm \frac{\pi}{6} \)

For \( n = 0 \): \( x = 2n\pi \pm \frac{\pi}{6} \implies x = \frac{\pi}{6} \)

For \( n = 1 \): \( x = 2\pi \pm \frac{\pi}{6} \implies x = \frac{11\pi}{6}, \frac{13\pi}{6} \)

\( \frac{13\pi}{6} > 2\pi \), so it is not included in the principal solution.

Principal solutions: \( x = \frac{\pi}{6} \) and \( \frac{11\pi}{6} \)
In simple words: For each equation, match it to a familiar angle using standard trigonometric values, then use the appropriate solution formula (which depends on the function type). Finally, plug in values of n and collect only those results that fall within [0, 2π).

Exam Tip: Always recall the standard solution formulas: for sine use \( n\pi + (-1)^n\alpha \), for cosine use \( 2n\pi \pm \alpha \), and for tangent use \( n\pi + \alpha \). Double-check that reciprocal identities are applied correctly (cosec, sec, cot to sin, cos, tan) before using the formula.

Question 2. Find the principal solutions of each of the following equations:
(i) \( \sin x = -\frac{1}{2} \)
(ii) \( \sqrt{2} \cos x + 1 = 0 \)
(iii) \( \tan x = -1 \)
(iv) \( \sqrt{3} \text{cosec } x + 2 = 0 \)
(v) \( \tan x = -\sqrt{3} \)
(vi) \( \sqrt{3} \sec x + 2 = 0 \)
Answer: To find the principal solution, isolate the trigonometric function, match it to a known angle, apply the appropriate solution formula, and filter for values in [0, 2π).

(i) Given: \( \sin x = -\frac{1}{2} \)

Using the formula \( \sin\theta = \sin\alpha \implies \theta = n\pi + (-1)^n\alpha \), we have

\( \sin x = -\frac{1}{2} = -\sin\frac{\pi}{6} = \sin(\pi + \frac{\pi}{6}) = \sin\frac{7\pi}{6} \implies x = n\pi + \frac{7\pi}{6}(-1)^n \)

For \( n = 0 \): \( x = 0 \times \pi + \frac{7\pi}{6}(-1)^0 \implies x = \frac{7\pi}{6} \)

For \( n = 1 \): \( x = 1 \times \pi + \frac{7\pi}{6}(-1)^1 \implies x = \pi - \frac{7\pi}{6} = -\frac{\pi}{6} \) [NOTE: \( -\frac{\pi}{6} = \frac{11\pi}{6} \)]

Principal solutions: \( x = \frac{7\pi}{6} \) and \( \frac{11\pi}{6} \)

(ii) Given: \( \sqrt{2} \cos x + 1 = 0 \implies \cos x = -\frac{1}{\sqrt{2}} \)

Using the formula \( \cos\theta = \cos\alpha \implies \theta = 2n\pi \pm \alpha \), we have

\( \cos x = -\frac{1}{\sqrt{2}} = \cos\frac{3\pi}{4} \implies x = 2n\pi \pm \alpha \)

For \( n = 0 \): \( x = 2 \times 0 \times \pi \pm \frac{3\pi}{4} \implies x = \frac{3\pi}{4} \)

For \( n = 1 \): \( x = 2\pi \pm \frac{3\pi}{4} \implies x = \frac{5\pi}{4}, \frac{11\pi}{4} \)

\( \frac{11\pi}{4} > 2\pi \), so it is not included in the principal solution.

Principal solutions: \( x = \frac{3\pi}{4} \) and \( \frac{5\pi}{4} \)

(iii) Given: \( \tan x = -1 \)

Using the formula \( \tan\theta = \tan\alpha \implies \theta = n\pi \pm \alpha \), we have

\( \tan x = -1 = \tan\frac{3\pi}{4} \implies x = n\pi + \alpha \)

For \( n = 0 \): \( x = n\pi + \frac{3\pi}{4} \implies x = \frac{3\pi}{4} \)

For \( n = 1 \): \( x = \pi + \frac{3\pi}{4} \implies x = \frac{7\pi}{4} \)

Principal solutions: \( x = \frac{3\pi}{4} \) and \( \frac{7\pi}{4} \)

(iv) Given: \( \sqrt{3} \text{cosec } x + 2 = 0 \implies \text{cosec } x = -\frac{2}{\sqrt{3}} \)

We know that \( \text{cosec}\theta \times \sin\theta = 1 \)

So \( \sin x = -\frac{\sqrt{3}}{2} \)

Using the formula \( \sin\theta = \sin\alpha \implies \theta = n\pi + (-1)^n\alpha \), we have

\( \sin x = -\frac{\sqrt{3}}{2} = \sin\frac{4\pi}{3} \implies \theta = n\pi + \frac{4\pi}{3}(-1)^n \)

For \( n = 0 \): \( x = 0 \times \pi + \frac{4\pi}{3}(-1)^0 \implies x = \frac{4\pi}{3} \)

For \( n = 1 \): \( x = 1 \times \pi + \frac{4\pi}{3}(-1)^1 \implies x = \pi - \frac{4\pi}{3} = -\frac{\pi}{3} \) [NOTE: \( -\frac{\pi}{3} = \frac{5\pi}{3} \)]

Principal solutions: \( x = \frac{4\pi}{3} \) and \( \frac{5\pi}{3} \)

(v) Given: \( \tan x = -\sqrt{3} \)

Using the formula \( \tan\theta = \tan\alpha \implies \theta = n\pi + \alpha \), we have

\( \tan x = -\sqrt{3} = \tan\frac{2\pi}{3} \implies x = n\pi + \alpha \)

For \( n = 0 \): \( x = n\pi + \frac{2\pi}{3} \implies x = \frac{2\pi}{3} \)

For \( n = 1 \): \( x = \pi + \frac{2\pi}{3} \implies x = \frac{5\pi}{3} \)

Principal solutions: \( x = \frac{2\pi}{3} \) and \( \frac{5\pi}{3} \)

(vi) Given: \( \sqrt{3} \sec x + 2 = 0 \implies \sec x = -\frac{2}{\sqrt{3}} \)

We know that \( \sec\theta \times \cos\theta = 1 \)

So \( \cos x = -\frac{\sqrt{3}}{2} \)

Using the formula \( \cos\theta = \cos\alpha \implies \theta = 2n\pi \pm \alpha \), we have

\( \cos x = -\frac{\sqrt{3}}{2} = \cos\frac{5\pi}{6} \implies x = 2n\pi \pm \frac{5\pi}{6} \)

For \( n = 0 \): \( x = 2 \times 0 \times \pi \pm \frac{5\pi}{6} \implies x = \frac{5\pi}{6} \)

For \( n = 1 \): \( x = 2\pi \pm \frac{5\pi}{6} \implies x = \frac{7\pi}{6}, \frac{17\pi}{6} \)

\( \frac{17\pi}{6} > 2\pi \), so it is not included in the principal solution.

Principal solutions: \( x = \frac{5\pi}{6} \) and \( \frac{7\pi}{6} \)
In simple words: Rearrange the equation to isolate the trigonometric function, recognize which quadrant contains the angle, then apply the standard formula for that function type to get all candidates. Keep only solutions within [0, 2π).

Exam Tip: When dealing with negative trigonometric values, identify the reference angle first, then place it in the correct quadrant(s). For composite equations like \( \sqrt{2}\cos x + 1 = 0 \), always isolate the trig function before applying solution formulas.

Question 3. Find the general solution of each of the following equations:
(i) \( \sin 3x = 0 \)
(ii) \( \sin\frac{3x}{2} = 0 \)
(iii) \( \sin\left(x + \frac{\pi}{5}\right) = 0 \)
(iv) \( \cos 2x = 0 \)
(v) \( \cos\frac{5x}{2} = 0 \)
(vi) \( \cos\left(x + \frac{\pi}{10}\right) = 0 \)
(vii) \( \tan 2x = 0 \)
(viii) \( \tan\left(3x + \frac{\pi}{6}\right) = 0 \)
(ix) \( \tan\left(2x - \frac{\pi}{4}\right) = 0 \)
Answer: We need to find the general solution for each equation. A solution that is generalized by means of periodicity is referred to as a general solution.

(i) Given: \( \sin 3x = 0 \)

Using the formula \( \sin\theta = 0 \implies \theta = n\pi, \, n \in \mathbb{I} \):

\( \sin 3x = 0 \implies 3x = n\pi \implies x = \frac{n\pi}{3} \, \text{where} \, n \in \mathbb{I} \)

So the general solution is \( x = \frac{n\pi}{3} \, \text{where} \, n \in \mathbb{I} \)

(ii) Given: \( \sin\frac{3x}{2} = 0 \)

Using the formula \( \sin\theta = 0 \implies \theta = n\pi, \, n \in \mathbb{I} \):

\( \sin\frac{3x}{2} = 0 \implies \frac{3x}{2} = n\pi \implies x = \frac{2n\pi}{3} \, \text{where} \, n \in \mathbb{I} \)

So the general solution is \( x = \frac{2n\pi}{3} \, \text{where} \, n \in \mathbb{I} \)

(iii) Given: \( \sin\left(x + \frac{\pi}{5}\right) = 0 \)

Using the formula \( \sin\theta = 0 \implies \theta = n\pi, \, n \in \mathbb{I} \):

\( \sin\left(x + \frac{\pi}{5}\right) = 0 \implies x + \frac{\pi}{5} = n\pi \implies x = n\pi - \frac{\pi}{5} \, \text{where} \, n \in \mathbb{I} \)

So the general solution is \( x = n\pi - \frac{\pi}{5} \, \text{where} \, n \in \mathbb{I} \)

(iv) Given: \( \cos 2x = 0 \)

Using the formula \( \cos\theta = 0 \implies \theta = (2n+1)\frac{\pi}{2}, \, n \in \mathbb{I} \):

\( \cos 2x = 0 \implies 2x = (2n+1)\frac{\pi}{2} \implies x = (2n+1)\frac{\pi}{4} \, \text{where} \, n \in \mathbb{I} \)

So the general solution is \( x = (2n+1)\frac{\pi}{4} \, \text{where} \, n \in \mathbb{I} \)

(v) Given: \( \cos\frac{5x}{2} = 0 \)

Using the formula \( \cos\theta = 0 \implies \theta = (2n+1)\frac{\pi}{2}, \, n \in \mathbb{I} \):

\( \cos\frac{5x}{2} = 0 \implies \frac{5x}{2} = (2n+1)\frac{\pi}{2} \implies x = (2n+1)\frac{\pi}{5} \, \text{where} \, n \in \mathbb{I} \)

So the general solution is \( x = (2n+1)\frac{\pi}{5} \, \text{where} \, n \in \mathbb{I} \)

(vi) Given: \( \cos\left(x + \frac{\pi}{10}\right) = 0 \)

Using the formula \( \cos\theta = 0 \implies \theta = (2n+1)\frac{\pi}{2}, \, n \in \mathbb{I} \):

\( \cos\left(x + \frac{\pi}{10}\right) = 0 \implies x + \frac{\pi}{10} = (2n+1)\frac{\pi}{2} \implies x = (2n+1)\frac{\pi}{2} - \frac{\pi}{10} \implies x = n\pi + \frac{2\pi}{5} \, \text{where} \, n \in \mathbb{I} \)

So the general solution is \( x = n\pi + \frac{2\pi}{5} \, \text{where} \, n \in \mathbb{I} \)

(vii) Given: \( \tan 2x = 0 \)

Using the formula \( \tan\theta = 0 \implies \theta = n\pi, \, n \in \mathbb{I} \):

\( \tan 2x = 0 \implies 2x = n\pi \implies x = \frac{n\pi}{2} \, \text{where} \, n \in \mathbb{I} \)

So the general solution is \( x = \frac{n\pi}{2} \, \text{where} \, n \in \mathbb{I} \)

(viii) Given: \( \tan\left(3x + \frac{\pi}{6}\right) = 0 \)

Using the formula \( \tan\theta = 0 \implies \theta = n\pi, \, n \in \mathbb{I} \):

\( \tan\left(3x + \frac{\pi}{6}\right) = 0 \implies 3x + \frac{\pi}{6} = n\pi \implies 3x = n\pi - \frac{\pi}{6} \implies x = \frac{n\pi}{3} - \frac{\pi}{18} \, \text{where} \, n \in \mathbb{I} \)

So the general solution is \( x = \frac{n\pi}{3} - \frac{\pi}{18} \, \text{where} \, n \in \mathbb{I} \)

(ix) Given: \( \tan\left(2x - \frac{\pi}{4}\right) = 0 \)

Using the formula \( \tan\theta = 0 \implies \theta = n\pi, \, n \in \mathbb{I} \):

\( \tan\left(2x - \frac{\pi}{4}\right) = 0 \implies 2x - \frac{\pi}{4} = n\pi \implies 2x = n\pi + \frac{\pi}{4} \implies x = \frac{n\pi}{2} + \frac{\pi}{8} \, \text{where} \, n \in \mathbb{I} \)

So the general solution is \( x = \frac{n\pi}{2} + \frac{\pi}{8} \, \text{where} \, n \in \mathbb{I} \)

Exam Tip: Identify which trigonometric function appears in the equation and apply the appropriate standard formula - different functions (sine, cosine, tangent) have distinct general solution patterns. Always isolate the angle before solving.

Question 4. Find the general solution of each of the following equations:
(i) \( \sin x = \frac{\sqrt{3}}{2} \)
(ii) \( \cos x = 1 \)
(iii) \( \sec x = \sqrt{2} \)
Answer: We need to find the general solution for each equation using the appropriate trigonometric formulas.

(i) Given: \( \sin x = \frac{\sqrt{3}}{2} \)

Using the formula \( \sin\theta = \sin\alpha \implies \theta = n\pi + (-1)^n\alpha, \, n \in \mathbb{I} \):

\( \sin x = \frac{\sqrt{3}}{2} = \sin\frac{\pi}{3} \implies x = n\pi + (-1)^n \cdot \frac{\pi}{3} \, \text{where} \, n \in \mathbb{I} \)

So the general solution is \( x = n\pi + (-1)^n \cdot \frac{\pi}{3} \, \text{where} \, n \in \mathbb{I} \)

(ii) Given: \( \cos x = 1 \)

Using the formula \( \cos\theta = \cos\alpha \implies \theta = 2n\pi \pm \alpha, \, n \in \mathbb{I} \):

\( \cos x = 1 = \cos(0°) \implies x = 2n\pi, \, n \in \mathbb{I} \)

So the general solution is \( x = 2n\pi \, \text{where} \, n \in \mathbb{I} \)

(iii) Given: \( \sec x = \sqrt{2} \)

We know that \( \sec\theta \times \cos\theta = 1 \), so \( \cos x = \frac{1}{\sqrt{2}} \)

Using the formula \( \cos\theta = \cos\alpha \implies \theta = 2n\pi \pm \alpha, \, n \in \mathbb{I} \):

\( \cos x = \frac{1}{\sqrt{2}} = \cos\frac{\pi}{4} \implies x = 2n\pi \pm \frac{\pi}{4}, \, n \in \mathbb{I} \)

So the general solution is \( x = 2n\pi \pm \frac{\pi}{4} \, \text{where} \, n \in \mathbb{I} \)

Exam Tip: First recognize what standard angle value the trigonometric function equals, then apply the corresponding general formula. For reciprocal functions like secant, convert to the basic function first.

Question 5. Find the general solution of each of the following equations:
(i) \( \cos x = \frac{-1}{2} \)
(ii) \( \csc x = -\sqrt{2} \)
(iii) \( \tan x = -1 \)
Answer: We need to find the general solution for each equation using the appropriate trigonometric formulas.

(i) Given: \( \cos x = \frac{-1}{2} \)

Using the formula \( \cos\theta = \cos\alpha \implies \theta = 2n\pi \pm \alpha, \, n \in \mathbb{I} \):

\( \cos x = \frac{-1}{2} = -\cos\frac{\pi}{3} = \cos\left(\pi - \frac{\pi}{3}\right) = \cos\frac{2\pi}{3} \implies x = 2n\pi \pm \frac{2\pi}{3}, \, n \in \mathbb{I} \)

So the general solution is \( x = 2n\pi \pm \frac{2\pi}{3} \, \text{where} \, n \in \mathbb{I} \)

(ii) Given: \( \csc x = -\sqrt{2} \)

We know that \( \csc\theta \times \sin\theta = 1 \), so \( \sin x = \frac{-1}{\sqrt{2}} \)

Using the formula \( \sin\theta = \sin\alpha \implies \theta = n\pi + (-1)^n\alpha, \, n \in \mathbb{I} \):

\( \sin x = \frac{-1}{\sqrt{2}} = \sin\frac{5\pi}{4} \implies x = n\pi + (-1)^n \cdot \frac{5\pi}{4} \, \text{where} \, n \in \mathbb{I} \)

So the general solution is \( x = n\pi + (-1)^n \cdot \frac{5\pi}{4} \, \text{where} \, n \in \mathbb{I} \)

(iii) Given: \( \tan x = -1 \)

Using the formula \( \tan\theta = \tan\alpha \implies \theta = n\pi + \alpha, \, n \in \mathbb{I} \):

\( \tan x = -1 = \tan\frac{3\pi}{4} \implies x = n\pi + \frac{3\pi}{4}, \, n \in \mathbb{I} \)

So the general solution is \( x = n\pi + \frac{3\pi}{4} \, \text{where} \, n \in \mathbb{I} \)

Exam Tip: For negative trigonometric values, identify the reference angle first, then determine which quadrant gives the required sign before applying the general formula.

Question 6. Find the general solution of each of the following equations:
(i) \( \sin 2x = \frac{1}{2} \)
(ii) \( \cos 3x = \frac{1}{\sqrt{2}} \)
(iii) \( \tan\frac{2x}{3} = \sqrt{3} \)
Answer: We need to find the general solution for each equation using the appropriate trigonometric formulas.

(i) Given: \( \sin 2x = \frac{1}{2} \)

Using the formula \( \sin\theta = \sin\alpha \implies \theta = n\pi + (-1)^n\alpha, \, n \in \mathbb{I} \):

\( \sin 2x = \frac{1}{2} = \sin\frac{\pi}{6} \implies 2x = n\pi + (-1)^n \cdot \frac{\pi}{6} \implies x = \frac{n\pi}{2} + (-1)^n \cdot \frac{\pi}{12}, \, n \in \mathbb{I} \)

So the general solution is \( x = \frac{n\pi}{2} + (-1)^n \cdot \frac{\pi}{12} \, \text{where} \, n \in \mathbb{I} \)

(ii) Given: \( \cos 3x = \frac{1}{\sqrt{2}} \)

Using the formula \( \cos\theta = \cos\alpha \implies \theta = 2n\pi \pm \alpha, \, n \in \mathbb{I} \):

Exam Tip: When the argument of the trigonometric function is not just \( x \), solve for that argument first, then divide by its coefficient to find \( x \). Always express the final answer with \( n \in \mathbb{I} \) clearly stated.

Question 7. Find the general solution of each of the following equations:
(i) \( \sec 3x = -2 \)
(ii) \( \cot 4x = -1 \)
(iii) \( \csc 3x = \frac{-2}{\sqrt{3}} \)
Answer: We need to find the general solution for each equation using the appropriate trigonometric formulas.

(i) Given: \( \sec 3x = -2 \)

We know that \( \sec\theta = \frac{1}{\cos\theta} \), so \( \cos 3x = \frac{1}{\sqrt{2}} \)

Using the formula \( \cos\theta = \cos\alpha \implies \theta = 2n\pi \pm \alpha, \, n \in \mathbb{I} \):

\( \cos 3x = \frac{1}{\sqrt{2}} = \cos\frac{\pi}{4} \implies 3x = 2n\pi \pm \frac{\pi}{4} \implies x = \frac{2n\pi}{3} \pm \frac{\pi}{12}, \, n \in \mathbb{I} \)

So the general solution is \( x = \frac{2n\pi}{3} \pm \frac{\pi}{12} \, \text{where} \, n \in \mathbb{I} \)

(ii) Given: \( \cot 4x = -1 \)

We know that \( \cot\theta = \frac{1}{\tan\theta} \), so \( \tan 4x = -1 \)

Using the formula \( \tan\theta = \tan\alpha \implies \theta = n\pi + \alpha, \, n \in \mathbb{I} \):

\( \tan 4x = -1 = \tan\left(-\frac{\pi}{4}\right) \implies 4x = n\pi - \frac{\pi}{4} \implies x = \frac{n\pi}{4} - \frac{\pi}{16}, \, n \in \mathbb{I} \)

So the general solution is \( x = \frac{n\pi}{4} - \frac{\pi}{16} \, \text{where} \, n \in \mathbb{I} \)

(iii) Given: \( \csc 3x = \frac{-2}{\sqrt{3}} \)

We know that \( \csc\theta = \frac{1}{\sin\theta} \), so we need to find what value of sine gives this result. This equation requires finding the reference angle and applying the appropriate formula for sine.

Exam Tip: For reciprocal trigonometric functions, always convert to the basic function first before applying the standard formula. Be careful with negative values and identify the correct reference angles.

Question 8. Find the general solution of each of the following equations:
(i) \( 4\cos^2 x = 1 \)
(ii) \( 4\sin^2 x - 3 = 0 \)
(iii) \( \tan^2 x = 1 \)
Answer:
(i) Given: \( 4\cos^2 x = 1 \Rightarrow \cos^2 x = \left(\frac{1}{4}\right) \)

\( \cos^2 x = \cos^2 \frac{\pi}{3} \)

Applying the formula \( \cos^2 \theta = \cos^2 \alpha \Rightarrow \theta = n\pi \pm \alpha, n \in I \):

\( x = n\pi \pm \frac{\pi}{3}, n \in I \)

Therefore, the general solution is \( x = n\pi \pm \frac{\pi}{3} \) where \( n \in I \)

(ii) Given: \( 4\sin^2 x - 3 = 0 \Rightarrow \sin^2 x = \frac{3}{4} = \sin^2 \frac{2\pi}{3} \)

\( \sin^2 x = \sin^2 \frac{2\pi}{3} \)

Using the formula \( \sin^2 \theta = \sin^2 \alpha \Rightarrow \theta = n\pi \pm \alpha, n \in I \):

\( x = n\pi \pm \frac{\pi}{3}, n \in I \)

Therefore, the general solution is \( x = n\pi \pm \frac{\pi}{3} \) where \( n \in I \)

(iii) Given: \( \tan^2 x = 1 \Rightarrow \tan^2 x = \tan^2 \frac{\pi}{4} \)

\( \tan^2 x = \tan^2 \frac{\pi}{4} \)

Applying the formula \( \tan^2 \theta = \tan^2 \alpha \Rightarrow \theta = n\pi \pm \alpha, n \in I \):

\( x = n\pi \pm \frac{\pi}{4}, n \in I \)

Therefore, the general solution is \( x = n\pi \pm \frac{\pi}{4} \) where \( n \in I \)
In simple words: When you have a squared trigonometric function, match it to a standard angle value, then apply the corresponding general formula. For squared sine and cosine, you get multiple solutions; for squared tangent, the approach is similar but follows its own pattern.

Exam Tip: Always recognize squared trig functions and match them to known angle values like \( \frac{\pi}{3}, \frac{\pi}{4}, \frac{\pi}{6} \). Use the correct general formula for each function to avoid missing solutions.

Question 9. Find the general solution of each of the following equations:
(i) \( \cos 3x = \cos 2x \)
(ii) \( \cos 5x = \sin 3x \)
(iii) \( \cos mx = \sin nx \)
Answer:
(i) Given: \( \cos 3x = \cos 2x \)

\( \cos 3x - \cos 2x = 0 \)

\( -2\sin\left(\frac{5x}{2}\right)\sin\left(\frac{x}{2}\right) = 0 \)

[NOTE: \( \cos C - \cos D = -2\sin\left(\frac{C+D}{2}\right)\sin\left(\frac{C-D}{2}\right) \)]

\( \sin\left(\frac{5x}{2}\right) = 0 \) or \( \sin\left(\frac{x}{2}\right) = 0 \)

Using the formula \( \sin\theta = 0 \Rightarrow \theta = n\pi, n \in I \):

\( \frac{5x}{2} = n\pi \) or \( \frac{x}{2} = m\pi \) where \( n, m \in I \)

\( x = \frac{2n\pi}{5} \) or \( x = 2m\pi \) where \( n, m \in I \)

Therefore, the general solution is \( x = \frac{2n\pi}{5} \) or \( x = 2m\pi \) where \( n, m \in I \)

(ii) Given: \( \cos 5x = \sin 3x \Rightarrow \cos 5x = \cos\left(\frac{\pi}{2} - 3x\right) \)

Using the formula \( \cos\theta = \cos\alpha \Rightarrow \theta = 2n\pi \pm \alpha, n \in I \):

\( 5x = 2n\pi + \left(\frac{\pi}{2} - 3x\right) \) or \( 5x = 2n\pi - \left(\frac{\pi}{2} - 3x\right) \)

\( 8x = 2n\pi + \frac{\pi}{2} \) or \( 2x = 2n\pi - \frac{\pi}{2} \)

\( x = \frac{n\pi}{4} + \frac{\pi}{16} \) or \( x = n\pi - \frac{\pi}{4} \) where \( n \in I \)

Therefore, the general solution is \( x = \frac{n\pi}{4} + \frac{\pi}{16} \) or \( x = n\pi - \frac{\pi}{4} \) where \( n \in I \)

(iii) Given: \( \cos mx = \sin nx \Rightarrow \cos mx = \cos\left(\frac{\pi}{2} - nx\right) \)

Using the formula \( \cos \mathfrak{a} = \cos \mathfrak{b} \Rightarrow \mathfrak{a} = 2k \mathfrak{b} \pm \mathfrak{b}, k \in I \):

\( mx = 2k\pi + \left(\frac{\pi}{2} - nx\right) \) or \( mx = 2k\pi - \left(\frac{\pi}{2} - nx\right) \)

(m+n)x = 2k\pi + \frac{\pi}{2} \) or \( (m-n)x = 2k\pi - \frac{\pi}{2} \)

\( x = \frac{2k\pi + \frac{\pi}{2}}{m+n} \) or \( x = \frac{2k\pi - \frac{\pi}{2}}{m-n} \) where \( k \in I \)
In simple words: When the angles inside cosine or sine are different, rearrange using sum-to-product formulas or convert sine to cosine. Then apply the standard formulas to break down the equation into simpler parts that you can solve.

Exam Tip: Remember to convert sine to cosine using \( \sin \theta = \cos\left(\frac{\pi}{2} - \theta\right) \) and always apply the product-to-sum formulas when you have sums or differences of trigonometric functions inside the equation.

Question 10. Find the general solution of each of the following equations:
\( \sin x = \tan x \)
Answer: Given: \( \sin x = \tan x \Rightarrow \sin x = \frac{\sin x}{\cos x} \)

\( \sin x \cos x = \sin x \)

\( \sin x \cos x - \sin x = 0 \)

\( \sin x (\cos x - 1) = 0 \)

\( \sin x = 0 \) or \( \cos x = 1 \)

Using the formulas \( \sin\theta = 0 \Rightarrow \theta = n\pi, n \in I \) and \( \cos\theta = \cos\alpha \Rightarrow \theta = 2k\pi \pm \alpha, k \in I \):

\( x = n\pi \) or \( x = 2k\pi \) where \( n, k \in I \)

Therefore, the general solution is \( x = n\pi \) or \( x = 2k\pi \) where \( n, k \in I \)
In simple words: Rewrite the equation by substituting the definition of tangent as sine over cosine. Multiply both sides by the denominator, then factor out the common sine term. This gives you two cases to solve separately.

Exam Tip: Always check your domain - tangent is undefined when cosine is zero, so verify that your solutions do not make the denominator vanish.

Question 11. Find the general solution of each of the following equations:
\( 4\sin x \cos x + 2\sin x + 2\cos x + 1 = 0 \)
Answer: Given: \( 4\sin x \cos x + 2\sin x + 2\cos x + 1 = 0 \)

\( 2\sin x (2\cos x + 1) + (2\cos x + 1) = 0 \)

\( (2\cos x + 1)(2\sin x + 1) = 0 \)

\( \cos x = -\frac{1}{2} = \cos\left(\frac{2\pi}{3}\right) \) or \( \sin x = -\frac{1}{2} = \sin\left(-\frac{\pi}{6}\right) \)

Applying the formulas \( \cos\theta = \cos\alpha \Rightarrow \theta = 2n\pi \pm \alpha \) or \( \sin\theta = \sin\alpha \Rightarrow \theta = m\pi + (-1)^m\alpha \) where \( n, m \in I \):

\( x = 2n\pi \pm \frac{2\pi}{3} \) or \( x = m\pi + (-1)^m \cdot \left(-\frac{\pi}{6}\right) \) where \( n, m \in I \)

Therefore, the general solution is \( x = 2n\pi \pm \frac{2\pi}{3} \) or \( x = m\pi + (-1)^m \cdot \left(-\frac{\pi}{6}\right) \) where \( n, m \in I \)
In simple words: Factor the equation by grouping - pull out the common factors from pairs of terms. Once you have two factors set to zero, you get two separate simpler equations to solve using standard formulas.

Exam Tip: When factoring trigonometric expressions, look for common factors and group terms strategically. Always double-check your factorization by expanding.

Question 12. Find the general solution of each of the following equations:
\( \sec^2 2x = 1 - \tan 2x \)
Answer: Given: \( \sec^2 2x = 1 - \tan 2x \)

\( 1 + \tan^2 2x = 1 - \tan 2x \)

\( \tan^2 2x + \tan 2x = 0 \)

\( \tan 2x (1 + \tan 2x) = 0 \)

\( \tan 2x = 0 \) or \( \tan 2x = -1 \)

Using the formulas \( \tan\theta = 0 \Rightarrow \theta = n\pi, n \in I \) and \( \tan\theta = \tan\alpha \Rightarrow \theta = k\pi \pm \alpha, k \in I \):

For \( \tan 2x = 0 \): \( 2x = n\pi \Rightarrow x = \frac{n\pi}{2} \) where \( n \in I \)

For \( \tan 2x = -1 = \tan\left(-\frac{\pi}{4}\right) \): \( 2x = k\pi - \frac{\pi}{4} \Rightarrow x = \frac{k\pi}{2} - \frac{\pi}{8} \) where \( k \in I \)

Therefore, the general solution is \( x = \frac{n\pi}{2} \) or \( x = \frac{k\pi}{2} - \frac{\pi}{8} \) where \( n, k \in I \)
In simple words: Use the identity that secant squared equals one plus tangent squared. Rearrange to get a quadratic-like equation in tangent, then factor out the common tangent term to find two cases.

Exam Tip: Always apply the Pythagorean identity \( \sec^2 \theta = 1 + \tan^2 \theta \) first when you see secant in the equation. This immediately converts the problem into a tangent-only equation.

Question 13. Find the general solution of each of the following equations:
\( \tan^3 x - 3\tan x = 0 \)
Answer: Given: \( \tan^3 x - 3\tan x = 0 \)

\( \tan x (\tan^2 x - 3) = 0 \)

\( \tan x = 0 \) or \( \tan x = \pm\sqrt{3} \)

\( \tan x = 0 \) or \( \tan x = \tan\left(\frac{\pi}{3}\right) \) or \( \tan x = \tan\left(-\frac{\pi}{3}\right) \)

Using the formulas \( \tan\theta = 0 \Rightarrow \theta = n\pi, n \in I \) and \( \tan\theta = \tan\alpha \Rightarrow \theta = k\pi \pm \alpha, k \in I \):

\( x = n\pi \) or \( x = k\pi + \frac{\pi}{3} \) or \( x = p\pi - \frac{\pi}{3} \) where \( n, k, p \in I \)

Therefore, the general solution is \( x = n\pi \) or \( x = k\pi + \frac{\pi}{3} \) or \( x = p\pi - \frac{\pi}{3} \) where \( n, k, p \in I \)
In simple words: Factor out the common tangent term. This gives you a cubic-like structure that breaks into one linear and one quadratic factor. Solve each piece using standard formulas.

Exam Tip: Recognize that \( \pm\sqrt{3} \) corresponds to \( \tan\left(\pm\frac{\pi}{3}\right) \). Always factor completely before applying the general solution formulas to avoid missing any solutions.

Question 14. Find the general solution of each of the following equations:
\( \sin x + \sin 3x + \sin 5x = 0 \)
Answer: Given: \( \sin x + \sin 3x + \sin 5x = 0 \)

\( (\sin 5x + \sin x) + \sin 3x = 0 \)

\( 2\sin 3x \cos 2x + \sin 3x = 0 \)

\( \sin 3x (2\cos 2x + 1) = 0 \)

[NOTE: \( \sin C + \sin D = 2\sin\frac{C+D}{2} \times \cos\frac{C-D}{2} \)]

\( \sin 3x = 0 \) or \( \cos 2x = -\frac{1}{2} = \cos\left(\frac{2\pi}{3}\right) \)

Using the formulas \( \sin\theta = 0 \Rightarrow \theta = n\pi, n \in I \) and \( \cos\theta = \cos\alpha \Rightarrow \theta = 2k\pi \pm \alpha, k \in I \):

\( 3x = n\pi \Rightarrow x = \frac{n\pi}{3} \) or \( 2x = 2k\pi \pm \frac{2\pi}{3} \Rightarrow x = k\pi \pm \frac{\pi}{3} \) where \( n, k \in I \)

Therefore, the general solution is \( x = \frac{n\pi}{3} \) or \( x = k\pi \pm \frac{\pi}{3} \) where \( n, k \in I \)
In simple words: Group the sum of sines strategically - combine the first and last terms using the sum-to-product formula. This creates a common factor that you can pull out to simplify the expression into two solvable pieces.

Exam Tip: Always look for sum-to-product formulas when you have sums of sines or cosines. Pairing terms strategically can reveal hidden factorizations that simplify the problem dramatically.

Question 15. Find the general solution of each of the following equations:
\( \sin x \tan x - 1 = \tan x - \sin x \)
Answer: Given: \( \sin x \tan x - 1 = \tan x - \sin x \)

\( \sin x \tan x + \sin x = \tan x + 1 \)

\( \sin x (\tan x + 1) = \tan x + 1 \)

Case 1: If \( \tan x + 1 \neq 0 \), then \( \sin x = 1 = \sin\left(\frac{\pi}{2}\right) \)

Case 2: If \( \tan x + 1 = 0 \), then \( \tan x = -1 = \tan\left(-\frac{\pi}{4}\right) \)

Using the formulas \( \sin\theta = \sin\alpha \Rightarrow \theta = n\pi + (-1)^n\alpha, n \in I \) and \( \tan\theta = \tan\alpha \Rightarrow \theta = k\pi \pm \alpha, k \in I \):

\( x = n\pi + (-1)^n \cdot \frac{\pi}{2} \) or \( x = k\pi - \frac{\pi}{4} \) where \( n, k \in I \)

Therefore, the general solution is \( x = n\pi + (-1)^n \cdot \frac{\pi}{2} \) or \( x = k\pi - \frac{\pi}{4} \) where \( n, k \in I \)
In simple words: Rearrange the equation to group terms with common factors. Factor out the bracketed expression, then handle the two cases - one where the factor equals zero and one where the other term must equal a specific value.

Exam Tip: When you factor and get an expression like \( A(B) = B \), be careful to split into two separate cases. Do not automatically cancel B unless you verify that B is never zero in your domain.

Question 16. Find the general solution of each of the following equations:
\( \cos x + \sin x = 1 \)
Answer: Given: \( \cos x + \sin x = 1 \)

\( \cos\left(x - \frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} = \cos\frac{\pi}{4} \)

[Divide \( \sqrt{2} \) on both sides and use \( \cos(x-y) = \cos x \cos y - \sin x \sin y \)]

Using the formula \( \cos\theta = \cos\alpha \Rightarrow \theta = 2k\pi \pm \alpha, k \in I \):

\( x - \frac{\pi}{4} = 2k\pi + \frac{\pi}{4} \) or \( x - \frac{\pi}{4} = 2k\pi - \frac{\pi}{4} \)

\( x = 2k\pi + \frac{\pi}{2} \) or \( x = 2k\pi \) where \( k \in I \)

Therefore, the general solution is \( x = 2n\pi + \frac{\pi}{2} \) or \( x = 2n\pi \) where \( n \in I \)
In simple words: Divide both sides by \( \sqrt{2} \) to rewrite the left side as a single cosine of a shifted angle. Then apply the standard cosine formula to find all solutions.

Exam Tip: When you have a sum of sine and cosine, divide by \( \sqrt{a^2 + b^2} \) (here \( \sqrt{2} \)) and use the angle addition formula to convert to a single trig function. This technique works for any linear combination of sine and cosine.

Question 17. Find the general solution of each of the following equations:
\( \cos x - \sin x = -1 \)
Answer: Given: \( \cos x - \sin x = -1 \)

\( \cos\left(x + \frac{\pi}{4}\right) = -\frac{1}{\sqrt{2}} = \cos\frac{3\pi}{4} \)

[Divide \( \sqrt{2} \) on both sides and use \( \cos(x-y) = \cos x \cos y - \sin x \sin y \)]

From \( \cos x - \sin x = -1 \), we also derive \( \sin x = 0 \) or \( \cos x = 0 \)

Using the formulas \( \cos\theta = \cos\alpha \Rightarrow \theta = 2k\pi \pm \alpha, k \in I \):

\( x + \frac{\pi}{4} = 2k\pi \pm \frac{3\pi}{4} \Rightarrow x = 2k\pi + \frac{3\pi}{4} - \frac{\pi}{4} \) or \( x = 2k\pi - \frac{3\pi}{4} - \frac{\pi}{4} \)

\( x = 2k\pi + \frac{\pi}{2} \) or \( x = 2k\pi - \pi \)

Therefore, the general solution is \( x = 2n\pi + \frac{\pi}{2} \) or \( x = (2n-1)\pi \) where \( n \in I \)
In simple words: Divide by \( \sqrt{2} \) and shift the angle by \( \frac{\pi}{4} \). Apply the cosine formula and simplify to get your final solutions.

Exam Tip: Pay attention to the sign - when cosine equals a negative value, use reference angles in the second and third quadrants. Verify your solutions by substituting back into the original equation.

Question 18. Find the general solution of each of the following equations:
\( \sqrt{\cos x + \sin x} = 1 \)
Answer: Given: \( \sqrt{\cos x + \sin x} = 1 \)

Squaring both sides: \( \cos x + \sin x = 1 \)

This is the same as Question 16. From the previous solution:

\( \cos\left(x - \frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} = \cos\frac{\pi}{4} \)

Using the formula \( \cos\theta = \cos\alpha \Rightarrow \theta = 2k\pi \pm \alpha, k \in I \):

\( x - \frac{\pi}{4} = 2k\pi \pm \frac{\pi}{4} \)

\( x = 2k\pi + \frac{\pi}{2} \) or \( x = 2k\pi \) where \( k \in I \)

However, we must verify that \( \cos x + \sin x \geq 0 \) (since it is under the square root). Both solutions satisfy this condition.

Therefore, the general solution is \( x = 2n\pi + \frac{\pi}{2} \) or \( x = 2n\pi \) where \( n \in I \)
In simple words: Square both sides to remove the square root, then solve the resulting equation the same way as Question 16. Always check that your final solutions make the expression inside the square root non-negative.

Exam Tip: When dealing with square roots of trigonometric expressions, always verify the domain after solving. Squaring can introduce extraneous solutions that do not satisfy the original equation.

Question 19. Find the general solution of each of the following equations:
\( 2\tan x - \cot x + 1 = 0 \)
Answer: Given: \( 2\tan x - \cot x + 1 = 0 \)

\( 2\tan x - \frac{1}{\tan x} + 1 = 0 \)

Multiplying through by \( \tan x \): \( 2\tan^2 x + \tan x - 1 = 0 \)

Factoring: \( (2\tan x - 1)(\tan x + 1) = 0 \)

\( \tan x = \frac{1}{2} \) or \( \tan x = -1 = \tan\left(-\frac{\pi}{4}\right) \)

For the first case \( \tan x = \frac{1}{2} \), let \( \alpha = \arctan\left(\frac{1}{2}\right) \). Then \( \tan x = \tan \alpha \Rightarrow x = k\pi + \alpha, k \in I \)

For the second case, using the formula \( \tan\theta = \tan\alpha \Rightarrow \theta = k\pi \pm \alpha, k \in I \):

\( x = k\pi - \frac{\pi}{4} \) where \( k \in I \)

Therefore, the general solution is \( x = k\pi + \arctan\left(\frac{1}{2}\right) \) or \( x = k\pi - \frac{\pi}{4} \) where \( k \in I \)
In simple words: Replace cotangent with one over tangent, then multiply through by tangent to convert to a polynomial equation in tangent. Factor this quadratic and solve each resulting equation separately using the standard tangent formula.

Exam Tip: When you multiply by \( \tan x \), remember that this can introduce solutions where \( \tan x \) is undefined. Always check that your solutions do not make tangent undefined (i.e., \( x \neq \frac{\pi}{2} + n\pi \)).

Question 20. Find the general solution of each of the following equations: sin x tan x – 1 = tan x – sin x
Answer: Starting with the given equation, we can rearrange it as sin x(tan x + 1) = tan x + 1. From this, we obtain either sin x = 1 = sin(π/2) or tan x = -1 = tan(3π/4).

Using the standard formulas, sin θ = sin α gives θ = nπ + (-1)ⁿα where n ∈ ℤ, and tan θ = tan α gives θ = kπ + α where k ∈ ℤ.

Therefore, the general solution is x = nπ + (-1)ⁿ(π/2) or x = kπ + (3π/4) where n, k ∈ ℤ.
In simple words: Rearrange the equation to factor out common terms, then solve each resulting equation separately. One case gives x = nπ + (-1)ⁿ(π/2) and the other gives x = kπ + (3π/4).

Exam Tip: Always factor and simplify trigonometric equations into basic forms before applying standard solution formulas - this step prevents algebraic errors.

Question 21. Find the general solution of each of the following equations: cot x + tan x = 2 cosec x
Answer: Beginning with the given equation, we convert to a common denominator: (cos²x + sin²x)/(sin x cos x) = 2 cosec x. This simplifies to 1 = sin 2x cosec x using the double-angle formula. Further simplification yields cosec 2x = cosec x, which leads to sin x = sin 2x. Expanding gives sin x = 2 sin x cos x, so either sin x = 0 or cos x = 1/2.

Using the standard formulas, sin θ = 0 gives θ = nπ where n ∈ ℤ, and cos θ = cos α gives θ = 2mπ ± α where m ∈ ℤ.

Therefore, the general solution is x = nπ or x = 2mπ ± (π/3) where n, m ∈ ℤ.
In simple words: Convert both sides to sine and cosine, use the identity sin²x + cos²x = 1, then solve the resulting basic equations. You get two families of solutions: one from sin x = 0 and another from cos x = 1/2.

Exam Tip: When simplifying trigonometric equations involving multiple functions, always convert to sine and cosine first and look for opportunities to apply fundamental identities.