Question

Solve for x :
cos x - cos 3x - sin 2x = 0.​

Answer 1

Answer:

The solutions are

$x=\frac{n\pi}{2},\:n\in\,Z$

$x=n\pi+(-1)^n\frac{\pi}{6},\:n\in\,Z$

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Step-by-step explanation:

Solve for x :

cos x - cos 3x - sin 2x = 0.​

$cosx-cos3x-sin2x=0$

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Using

$\boxed{cosC-cosD=-2\,sin(\frac{C+D}{2})\,sin(\frac{C-D}{2})}$

$-2\,sin(\frac{x+3x}{2})\,sin(\frac{x-3x}{2})-sin2x=0$

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$-2\,sin2x\,sin(-x)-sin2x=0$

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$2\,sin2x\,sinx-sin2x=0$

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$sin2x(2sinx-1)=0$

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$sin2x=0\:(or)\:2sinx-1=0$

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$\text{ case(i): }sin2x=0$

$2x=n\pi,\:n\in\,Z$

$\implies\:\boxed{x=\frac{n\pi}{2},\:n\in\,Z}$

$\text{case(ii):}2sinx-1=0$

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$sinx=\frac{1}{2}$

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$sinx=sin\frac{\pi}{6}$

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$\implies\,\boxed{x=n\pi+(-1)^n\frac{\pi}{6}\:n\in\,Z}$

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