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!
 A: 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? 
A: Observe:
$$
\cos x-\sin x+2\sin x\cos x+6\cos^2 x-2=(\sin x)(2\cos x-1)+(2\cos x-1)(3\cos x+2)=(\sin x+3\cos x-2)(2\cos x-1).
$$
Therefore, $\cos x=\frac{1}{2}$ is a solution.
The solutions to $\sin x+3\cos x+2=0$ are less pleasant, but one can solve $\sin x=-2-3\cos x$, square both sides and solve, using the quadratic formula for $\cos x$.
A: I pick the solution by Michael Burr and Zain Patel and focus on the roots of $f(x)=3\cos x + \sin x +2$:
using this page you may write:
\begin{align}
f(x)&=\sqrt{10}\sin\Big(x+\arcsin\frac{3}{\sqrt{10}}\Big)+2
\end{align}
hence the roots are $x_k=-\arcsin\frac{3}{\sqrt{10}}-\arcsin\frac{\sqrt 2}{\sqrt{5}}+2k\pi$
A: 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 $\cos x=((6\pm \sqrt{6})/(10))$. 
