Solve: $3\sin{2x}+4\cos{2x}-2\cos{x}+6\sin{x}-6=0$ 
Solve:   $3\sin{2x}+4\cos{2x}-2\cos{x}+6\sin{x}-6=0$

My Try
$6\sin{x}\cos{x}+4(\cos^2{x}-\sin^2{x})-2\cos{x}+6\sin{x}-6=0$
I have expanded the equation, But I cannot proceed further, Any hint would be appreciated. Thank you!

Abbreviating $\sin x$ and $\cos x$ to $s$ and $c$, and writing $\sin2x=2sc$ and $\cos2x=2c^2-1$, the equation becomes
$$6sc+4(2c^2-1)-2c+6s-6=0$$
Factoring out a $2$ and grouping terms, we rewrite this as
$$3s(c+1)+4c^2-c-5=0$$
But $4c^2-c-5=(4c-5)(c+1)$, so we have two solutions in $s$ and $c$:
$$c=-1\quad\text{and}\quad3s+4c=5$$
Reverting to $\sin x$ and $\cos x$ and recognizing the Pythagorean triple $3^2+4^2=5^2$, these become
$$\cos x=-1\quad\text{and}\quad\sin(x+\theta)=1$$
where $\sin\theta=4/5$ and $\cos\theta=3/5$, e.g. $\theta=\arcsin(4/5)$. So the solution set is
$$\{\pi+2n\pi\mid n\in\mathbb{Z}\}\cup\{\pi/2-\arcsin(4/5)+2n\pi\mid n\in\mathbb{Z} \}$$
Remark: The abbreviations $s$ and $c$ are merely a convenience. Once I get all the sines and cosines down to $\sin x$ and $\cos x$, I find it tedious to write the functions out in full over and over again.
