# 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)

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