# How to Find General Solution of Trigonometric Equation

Here, you will learn what is trigonometric equation and how to find general solution of trigonometric equation with examples.

Let’s begin –

An equation involving one or more trigonometrical ratios of unknown angles is called a trigonometrical equation.

## Solution of Trigonometric Equation

A value of the unknown angle which satisfies the given equation is called a solution of the trigonometric equation.

(a) Principal solution :- The solution of the trigonometric equation lying in the interval [0, $$2\pi$$).

(b) General Solution :- Since all the trigonometric functions are many one & periodic, hence there are infinite values values of 0 for which trigonometric functions have the same value. All such possible values of 0 for which the given trigonometric function is satisfied is given by a general formula. Such a general formula is called general solution of trigonometric equation.

(c) Particular Solution :- The solution of the trigonometric equation lying in the given interval.

## General Solution of Trigonometric Equation

(a) If sin $$\theta$$ = 0, then $$\theta$$ = n$$\pi$$, n $$\in$$ I (set of integers)

(b) If cos $$\theta$$ = 0, then $$\theta$$ = (2n+1)$$\pi\over 2$$, n $$\in$$ I

(c) If tan $$\theta$$ = 0, then $$\theta$$ = n$$\pi$$, n $$\in$$ I

Example : Find the general solution of trigonometric equation :
(i) $$sin2\theta$$ = 0
(ii) $$tan{3\theta\over 4}$$ = 0

Solution :
(i) $$sin2\theta$$ = 0

$$\implies$$ $$2\theta$$ = $$n\pi$$, where n $$\in$$ Z     [sin $$\theta$$ = 0, then $$\theta$$ = n$$\pi$$]

$$\implies$$ $$\theta$$ = $$n\pi\over 2$$, where n $$\in$$ Z

(ii) $$tan{3\theta\over 4}$$ = 0

$$\implies$$ $$3\theta\over 4$$ = $$n\pi$$, where n $$\in$$ Z     [tan $$\theta$$ = 0, then $$\theta$$ = n$$\pi$$]

$$\implies$$ $$\theta$$ = $$4n\pi\over 3$$, where n $$\in$$ Z

(d) If sin $$\theta$$ = sin $$\alpha$$, then $$\theta$$ = n$$\pi$$ + $${(-1)}^n\alpha$$, where $$\alpha$$ $$\in$$ [-$$\pi\over 2$$, $$\pi\over 2$$], n $$\in$$ I

(e) cos $$\theta$$ = cos $$\alpha$$, then $$\theta$$ = 2n$$\pi$$ $$\pm$$ $$\alpha$$, where $$\alpha$$ $$\in$$ [0,$$\pi$$], n $$\in$$ I

(f) tan $$\theta$$ = tan $$\alpha$$, then $$\theta$$ = n$$\pi$$ + $$\alpha$$, where $$\alpha$$ $$\in$$ (-$$\pi\over 2$$, $$\pi\over 2$$), n $$\in$$ I

Example : Find the general solution of trigonometric equation :
(i) $$sin\theta$$ = $$\sqrt{3}\over 2$$
(ii) $$cos3\theta$$ = $$-1\over 2$$

Solution :
(i) A value of $$\theta$$ satisfying $$sin\theta$$ = $$\sqrt{3}\over 2$$ is $$\pi\over 3$$

$$sin\theta$$ = $$\sqrt{3}\over 2$$

$$\implies$$ $$sin\theta$$ = $$sin\pi\over 3$$ $$\implies$$ $$\theta$$ = n$$\pi$$ + $${(-1)}^n{\pi\over 3}$$,

(ii) $$cos3\theta$$ = $$-1\over 2$$

$$\implies$$ $$cos3\theta$$ = $$cos{2\pi\over 3}$$ $$\implies$$ $$3\theta$$ = 2n$$\pi$$ $$\pm$$ $$2\pi\over 3$$

$$\implies$$ $$\theta$$ = 2n$$\pi\over 3$$ $$\pm$$ $$2\pi\over 9$$

(g) If sin $$\theta$$ = 1, then $$\theta$$ = 2n$$\pi$$ + $$\pi\over 2$$ = (4n + 1)$$\pi\over 2$$, n $$\in$$ I

(h) If cos $$\theta$$ = 1, then $$\theta$$ = 2n$$\pi$$, n $$\in$$ I

(i) If $$sin^2\theta$$ = $$sin^2\alpha$$ or $$cos^2\theta$$ = $$cos^2\alpha$$ or $$tan^2\theta$$ = $$tan^2\alpha$$, then $$\theta$$ = n$$\pi$$ $$\pm$$ $$\alpha$$, n $$\in$$ I

Example : Find the general solution of trigonometric equation $$7cos^2\theta$$ + $$3sin^2\theta$$ = 4

Solution : We have,

$$7cos^2\theta$$ + $$3sin^2\theta$$ = 4

$$7(1-sin^2\theta)$$ + $$3sin^2\theta$$ = 4

$$4sin^2\theta$$ = 3

$$sin^2\theta$$ = $$3\over 4$$ = $$({\sqrt{3}\over 2})^2$$

$$sin^2\theta$$ = $$sin^2{\pi\over 3}$$ $$\implies$$ $$\theta$$ = $$n\pi\pm{\pi\over 3}$$.

Hope you learnt how to solve general solution of trigonometric equation, learn more concepts of trigonometric equation and practice more questions to get ahead in the competition. Good luck!