First of all, we can use the double angle identity to write
[tex]\cos(2x) = 2\cos^2(x)-1[/tex]
The equation becomes
[tex]2\cos^2(x)-1+\cos^2(x) = 1 \iff 3\cos^2(x) = 2[/tex]
Divide both sides by 3 to get
[tex]\cos^2(x) = \dfrac{2}{3}[/tex]
And finally consider the square root of both terms (don't forget the double sign):
[tex]\cos(x) = \pm\sqrt{\dfrac{2}{3}}[/tex]
So, the solutions are
[tex]x = \arccos\left(\sqrt{\dfrac{2}{3}}\right)\approx 35^\circ\\x = -\arccos\left(\sqrt{\dfrac{2}{3}}\right)\approx -35^\circ[/tex]
If you want x in [0,360], consider the equivalent angle
[tex]-35+360 = 325[/tex]