Question

I don't have a clue please help

Answer 1

First of all, we can use the double angle identity to write

$\cos(2x) = 2\cos^2(x)-1$

The equation becomes

$2\cos^2(x)-1+\cos^2(x) = 1 \iff 3\cos^2(x) = 2$

Divide both sides by 3 to get

$\cos^2(x) = \dfrac{2}{3}$

And finally consider the square root of both terms (don't forget the double sign):

$\cos(x) = \pm\sqrt{\dfrac{2}{3}}$

So, the solutions are

$x = \arccos\left(\sqrt{\dfrac{2}{3}}\right)\approx 35^\circ\\x = -\arccos\left(\sqrt{\dfrac{2}{3}}\right)\approx -35^\circ$

If you want x in [0,360], consider the equivalent angle

$-35+360 = 325$