Solving $2\cos2x-4\sin x\cos x=\sqrt{6}$ 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?

Note that
$$2\cos(2x)-2\sin (2x)=\sqrt{2^2+2^2}\cos(2x+\alpha)$$
where
$$\cos\alpha=\sin\alpha=\frac{2}{2\sqrt 2}=\frac{1}{\sqrt 2}.$$
Hence, you'll have
$$2\sqrt 2\cos(2x+45^\circ)=\sqrt 6\Rightarrow\cos(2x+45^\circ)=\frac{\sqrt 6}{2\sqrt 2}=\frac{\sqrt 3}{2}.$$
