# How do you solve Cos(2x)cos(x)-sin(2x)sin(x)=0?

May 27, 2018

We have:

$\cos \left(2 x\right) \cos x = \sin \left(2 x\right) \sin x$

$1 = \tan \left(2 x\right) \tan x$

Now we have to make some manipulations using $\tan \left(2 x\right) = \frac{2 \sin x \cos x}{{\cos}^{2} x - {\sin}^{2} x}$

$1 = \frac{2 \sin x \cos x}{{\cos}^{2} x - {\sin}^{2} x} \cdot \sin \frac{x}{\cos} x$

${\cos}^{2} x - {\sin}^{2} x = 2 {\sin}^{2} x$

$1 - {\sin}^{2} x - {\sin}^{2} x = 2 {\sin}^{2} x$

$1 = 4 {\sin}^{2} x$

$\frac{1}{4} = {\sin}^{2} x$

$\sin x = \pm \frac{1}{2}$

It's now clear that $x = \frac{\pi}{6} , \frac{5 \pi}{6} , \frac{7 \pi}{6} \mathmr{and} \frac{11 \pi}{6}$. In general form, these are $\frac{\pi}{6} + \pi n$ and $\frac{5 \pi}{6} + \pi n$. However we must also include when $\cos x = 0$, namely when $x = \frac{\pi}{2} + \pi n$. We can confirm graphically.

Hopefully this helps!