Giải phương trình 9sinx+6cosx−3sin2x+cos2x=8.

\(9\sin x + 6\cos x - 3\sin 2x + \cos 2x = 8\)

\(9\sin x + 6\cos x - 6\sin x\cos x + 1 - 2{\sin ^2}x = 8\)

\(9\sin x + 6\cos x\left( {1 - \sin x} \right) - 2{\sin ^2}x - 7 = 0\)

\(6\cos x\left( {1 - \sin x} \right) + \left( {1 - \sin x} \right)\left( {2 - \sin 7x} \right) = 0\)

\(\left( {1 - \sin x} \right)\left( {6\cos x + 2 - \sin 7x} \right) = 0\)

\[ \Rightarrow x = \frac{\pi }{2} + 2k\pi \]