Q:

$ \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?

A:

Linearise the left-hand side: we start from formula:
$$\cos^2u=\frac12(1+\cos 2u)$$
from which we obtain the equation:
$$\frac12(3+\cos 2x+\cos 4x+\cos 6x)=\frac32\iff\cos 2x+\cos 4x+\cos 6x=0.$$
Now factorise the left-hand side:
\begin{align*}
\cos 2x+\cos 4x+\cos 6x&=\cos(4x- 2x)+\cos 4x+\cos(4x+2x)\\
&=2\cos4x\cos2x+\cos 4x=\cos 4x(2\cos2x+1)
\end{align*}
Thus the equation is equivalent to the system of standard equations:
$$\begin{cases}
\cos 4x=0\\\cos2x=-\dfrac12\end{cases}$$
Solutions:
$$\begin{cases}
4x\equiv \dfrac\pi2&\bmod\pi\\
2x\equiv \pm\dfrac{2\pi}3&\bmod2\pi
\end{cases}\iff \begin{cases}
x\equiv \dfrac\pi8&\bmod\dfrac\pi4\\
x\equiv \pm\dfrac\pi3&\bmod\pi
\end{cases}$$

