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#
