<s>[INST] $ \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?
 [/INST] I shall use $\sum \cos$ when angles are in arithmetic progression  whose usefulness is more apparent if the number of summands is more than three.
If $\sin x=0,x=n\pi\implies\cos2x+\cos4x+\cos6x=1+1+1\ne0$
So, $\cos2x+\cos4x+\cos6x=0\implies\sin x\ne0$
$$\dfrac{2\sin x\cos2x+2\sin x\cos4x+2\sin x\cos6x}{2\sin x}=\dfrac{\sin7x-\sin x}{2\sin x}$$
Now $\sin7x=\sin x\implies7x=m\pi+(-1)^mx$ where $m$ is any integer
If $m$ is even $=2r$(say) $7x=2r\pi+x\iff x=\dfrac{r\pi}3$
If $m$ is odd $=2r+1$(say)  $7x=(2r+1)\pi-x\iff x=\dfrac{(2r+1)\pi}8$
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