# Trigonometric Equation Examples

Here you will learn some trigonometric equation examples for better understanding of trigonometric equation concepts.

Example 1 : Find general solution of (2sinx – cosx)(1 + cosx) = $$sin^2x$$

Solution : (2sinx – cosx)(1 + cosx) – (1 – $$cos^2x$$) = 0

$$\therefore$$   (1 + cosx)(2sinx – cosx – 1 + cosx) = 0

$$\therefore$$   (1 + cosx)(2sinx – 1) = 0

$$\implies$$   cosx = -1   or   sinx = $$1\over 2$$

$$\implies$$   cosx = -1 = cos$$\pi$$   $$\implies$$   x = 2n$$\pi$$ + $$\pi$$ = (2n+1)$$\pi$$, n $$\in$$ I

or   sinx = $$1\over 2$$ = sin$$\pi\over 6$$   $$\implies$$   x = k$$\pi + (-1)^k{\pi\over 6}$$, k $$\in$$ I

Example 2 : Solve : 6 – 10cosx = 3$$sin^2x$$

Solution : we have, 6 – 10cosx = 3$$sin^2x$$

$$\therefore$$   6 – 10cosx = 3 – 3$$cos^2x$$

$$\implies$$   3$$cos^2x$$ – 10cosx + 3 = 0

$$\implies$$   (3cosx-1)(cosx-3) = 0   $$\implies$$   cosx = $$1\over 3$$ or cosx = 3

Since cosx =3 is not possible as -1 $$\le$$ cosx $$\le$$ 1

$$\therefore$$   cosx = $$1\over 3$$ = cos($$cos^{-1}{1\over 3}$$)   $$\implies$$   x = 2n$$\pi$$ $$\pm$$ $$cos^{-1}{1\over 3}$$

Example 3 : Solve : cos3x + sin2x – sin4x = 0

Solution : we have, cos3x + (sin2x – sin4x) = 0

= cos3x – 2sinx.cos3x = 0

$$\implies$$   (cos3x)(1 – 2sinx) = 0

$$\implies$$   cos3x = 0   or   sinx = $$1\over 2$$

$$\implies$$   cos3x = 0 = cos$$\pi\over 2$$   or   sinx = $$1\over 2$$ = sin$$\pi\over 6$$

$$\implies$$   3x = 2n$$\pi$$ $$\pm$$ $$\pi\over 2$$   or   x = m$$\pi$$ + $${(-1)}^m$$$$\pi\over 6$$

$$\implies$$   x = $$2n\pi\over 3$$ $$\pm$$ $$\pi\over 6$$   or   x = m$$\pi$$ + $${(-1)}^m$$$$\pi\over 6$$; (n, m $$\in$$ I)

Practice these given trigonometric equation examples to test your knowledge on concepts of trigonometric equation.