cos 3x sin2x = 1 /16 (2 cos x - cos 3x - cos 5x)

**blue_blue** · April 26th 2009, 08:40 AM

Please solve cos 3x sin2x = 1 /16 (2 cos x - cos 3x - cos 5x)

**red_dog** · April 27th 2009, 07:56 AM

$16\cos 3x\sin 2x=(\cos x-\cos 3x)+(\cos x-\cos 5x)$

$16\cos 3x\sin 2x=2\sin x\sin 2x+2\sin 2x\sin 3x$

$\sin 2x(8\cos 3x-\sin x-\sin 3x)=0$

$\sin 2x=0\Rightarrow 2x=k\pi\Rightarrow x=\frac{k\pi}{2}, \ k\in\mathbf{Z}$

$8\cos 3x-(\sin x+\sin 3x)=0$

$8\cos 3x-2\sin 2x\cos x=0$

$4(4\cos^3x-3\cos x)-\sin 2x\cos x=0$

$\cos x(16\cos^2x-12-\sin 2x)=0$

$\cos x=0\Rightarrow x=\frac{(2k+1)\pi}{2}, \ k\in\mathbf{Z}$

$16\cos^2x-12-2\sin x\cos x=0$

$16\cos^2x-12(\sin^2x+\cos^2x)-2\sin x\cos x=0$

$2\cos^2x-6\sin^2x-\sin x\cos x=0$

Divide by $\sin^2x$

$2-6\tan^2x-\tan x=0$

Substitute $\tan x=t$ and solve the quadratic.

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