<s>[INST] 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!
 [/INST] Let $t = \tan (x/2)$.  Then $\sin x = 2t/(1+t^2)$ and $\cos x = (1-t^2)/(1+t^2)$.  Your equation looks quartic after multiplying both sides by $(1+t^2)^2$.
$$12t(1-t^2)+4((1-t^2)^2-4t^2)-(2-2t^2-12t)(1+t^2)-6(1+t^2)^2$$
but it simplifies to $-4(3t-1)^2=0$, so $t = 1/3$, giving $x = 2(\tan^{-1} t + n\pi) = 2(\tan^{-1}{\frac13}+n\pi)$ for any integer $n$, since $\tan$ has period $\pi$.

The above $t$-substitution omits the possibility that $x = (2n+1) \pi$, which is actually a solution: cross out the $\sin$ in the equation and observe that $4(1)-2(-1)-6 = 0$.
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