Find the principal solutions of the trigonometric equation $\cos x-\sin x+\sin 2x+3\cos2x+1=0$ I am unable to simplify the expression. If I simplify the double angles, it leaves me with a nasty expression, 
$\cos x-\sin x+2\sin x\cos x+6\cos^2 x-2=0$. What do I do next. Some hints, please. Also, is there some elegant solution? Thanks!
Hint: Let $a = \cos x$ and $b = \sin x$ so that you get $$a - b + 2ab + 6a^2 - 2 = 0 \iff (2a-1)(3a + b + 2) = 0$$

One solution is already apparent, $\cos x = \frac{1}{2}$, can you work out the others?
You have $\cos x-\sin x+\sin 2x+3\cos 2x+1=0$. Or $\sin x-\sin 2x=\cos x+3\cos 2x+1$. Or $\sin x(1-2\cos x)=\cos x+6\cos^2x-2$.

Either $\cos x =1/2$ or $\sin x=((6\cos^2x+\cos x-2)/(1-2\cos x))$. Or $\sin x=-3\cos x-2$. Squaring we get $10\cos^2x-12\cos x+3=0$. Thus $\cos x=((6\pm \sqrt{6})/(10))$.