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- Feb 14, 2012

- 3,909

Without thinking much, I do the following:

$2cos3xcos2x=cos6x$

Now, even if I still refuse to devise a plan but blindly applying the trig formula, I get:

$2cos3xcos2x=cos^23x-sin^23x$

Hmm, clearly this isn't going to bring me to any credible output.

But questions like this often yield the cases where:

1. zero=zero

2. 1 = 1

Thus, without actually solving it, I can say that

$x=\frac{\pi}{4}, \frac{\pi}{3}, and \frac{3\pi}{4}$ in the interval $0<x<2\pi$. I must have left out some other values. That is for sure.

That is my thought.

Despite my having said this, I tried to expand it and get down to all terms involving only $cos^nx$ and/or $sin^nx$

$2cos3xcos2x=cos^23x-sin^23x$

$2(4cos^3x-3cosx)(2cos^2x-1)=(4cos^3x-3cosx)^2-(3sinx-4sin^3x)^2$

Ah! This is a move in the wrong direction.

Could someone help me, please?

Thanks.