How I solve the following equation for $0 \le x \le 360$:
$$ 2\cos2x-4\sin x\cos x=\sqrt{6} $$
I tried different methods. The first was to get things in the form of $R\cos(x \mp \alpha)$:
$$ 2\cos2x-2(2\sin x\cos x)=\sqrt{6}\\ 2\cos2x-2\sin2x=\sqrt{6}\\ R = \sqrt{4} = 2 \\ \alpha = \arctan \frac{2}{2} = 45\\ \therefore \cos(2x + 45) = \frac{\sqrt6}{2} $$
which is impossible. I then tried to use t-substitution, where:
$$ t = \tan\frac{x}{2}, \sin x=\frac{2t}{1+t^2}, \cos x =\frac{1-t^2}{1+t^2} $$
but the algebra got unreasonably complicated. What am I missing?