Giải phương trình 1+sinx+cos3x=cosx+sin2x+cos2x.
Giải thích
\(1 + \sin x + \cos 3x = \cos x + \sin 2x + \cos 2x\)
\(\left( {1 - \cos 2x} \right) + \left( {\cos 3x - \cos x} \right) + \sin x - \sin 2x = 0\)
\(2{\sin ^2}x - 2\sin 2x\sin x + \sin x - 2\sin x\cos x = 0\)
\(\sin x\left( {2\sin x - 2\sin 2x + 1 - 2\cos x} \right) = 0\)
\(\sin x\left[ {2\sin x\left( {1 - \cos 2x} \right) + \left( {1 - 2\cos x} \right)} \right] = 0\)
\(\sin x\left( {1 - 2\cos x} \right)\left( {2\sin x + 1} \right) = 0\)
\[ \Rightarrow x \in \left\{ {k\pi ; \pm \frac{\pi }{3} + 2k\pi ; - \frac{\pi }{6} + 2k\pi ;\frac{{7\pi }}{6} + 2k\pi } \right\}\]