Given that
\cos^2\theta=2\cos^2\frac{\theta}{2}
(2\cos^2\frac{\theta}{2}-1)^2=2\cos^2\frac{\theta}{2}
4\cos^4\frac{\theta}{2}+1-4\cos^2\frac{\theta}{2}=2\cos^2\frac{\theta}{2}
4\cos^4\frac{\theta}{2}-6\cos^2\frac{\theta}{2}+1=0
Solving above quadratic equation in terms of \cos^2\frac{\theta}{2} using quadratic formula as follows
\cos^2\frac{\theta}{2}=\frac{-(-6)\pm\sqrt{(-6)^2-4(4)(1)}}{2(4)}
\cos^2\frac{\theta}{2}=\frac{3\pm\sqrt{5}}{4}
But , 0\le\cos^2\frac{\theta}{2}\le1
\therefore \cos^2\frac{\theta}{2}=\frac{3-\sqrt{5}}{4}
\cos^2\frac{\theta}{2}=(\frac{\sqrt5-1}{2\sqrt2})^2
Let \cos^2\frac{\theta}{2}=\cos^2\alpha
\frac{\theta}{2}=n\pi\pm\alpha
\theta=2(n\pi\pm\alpha)
Where, \cos\alpha=\frac{\sqrt5-1}{2\sqrt2}=64.09^\circ &
n is any integer i.e. n=0, \pm1, \pm2, \pm3, \ldots
But since, 0\le\theta\le360^\circ hence, setting n=0\ &\ 1 & taking positive & negative signs respectively, we get two values of \theta as follows
\theta=2(0+64.09^\circ), 2(180^\circ-64.09^\circ)
=128.18^\circ, 231.82^\circ