Giải phương trình: sinx+sin2x+sin3x=1+cosx+cos2x.

\[\sin x + \sin 2x + \sin 3x = 1 + \cos x + \cos 2x\]

\[\left( {\sin x + \sin 3x} \right) + \sin 2x = \left( {1 + \cos 2x} \right) + \cos x\]

\[2\sin 2x\cos x + \sin 2x = 2{\cos ^2}x + \cos x\]

\[\sin 2x\left( {2\cos x + 1} \right) = \cos x\left( {2\cos x + 1} \right)\]

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

\[x \in \left\{ {\frac{\pi }{2} + k\pi \,;\,\, \pm \frac{{2\pi }}{3} + 2k\pi \,;\,\,\frac{\pi }{6} + 2k\pi \,;\,\,\frac{\pi }{6} + 2k\pi } \right\}\].