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?

You have made a mistake in the third step.
$$
2\cos2x-2(2\sin x\cos x)=\sqrt{6}\\
2\cos2x-2\sin2x=\sqrt{6}\\
R = \sqrt{8} = 2\sqrt{2} \\
\alpha = \arctan -\frac{2}{2} = -\frac{\pi}{4}\\
\therefore \cos(2x - \frac{\pi}{4}) = \frac{\sqrt3}{2}
$$
Hope this help.
