$ \cos ^2\left(x\right)+\cos ^2\left(2x\right)+\cos ^2\left(3x\right)=\frac{3}{2} $ $$ \cos ^2\left(x\right)+\cos ^2\left(2x\right)+\cos ^2\left(3x\right)=\frac{3}{2} $$
How can I solve this one, I mean I get something like this:
$-3+\left(-1+2\cos ^2\left(x\right)\right)^22+2\left(-3\cos \left(x\right)+4\cos ^3\left(x\right)\right)^2+2\cos ^2\left(x\right)=0$
This equation seems rather hard to solve from here, any tips or other ways to come to an solution?

Let $\mu = \cos 2x$ for shortness.
Squaring
$$
\cos (2x\pm x) = \cos 2x \cos x \mp \sin 2x \sin x
$$
we obtain
$$
\cos^2 (2x\pm x) = \cos^2 2x \cos^2 x + \sin^2 2x \sin^2 x \mp \text{cancelling terms} = \\
= \mu^2 \frac{1+\mu}{2}+(1-\mu^2)\frac{1-\mu}{2} \mp \text{cancelling terms} = \\ =
\mu^3 - \frac{\mu}{2} + \frac{1}{2} \mp \text{cancelling terms}
$$
So the original equation becomes
$$
2\mu^3 + \mu^2 - \mu + 1 = \frac{3}{2}\\
2\mu^3 + \mu^2 - \mu - \frac{1}{2} = 0\\
(2\mu+1)(2\mu^2 - 1) = 0
$$
with solutions
$$
\mu_1 = -\frac{1}{2},\quad \mu_{2,3} = \pm \frac{1}{\sqrt{2}}
$$
