Giải phương trình 1 + sin x + cos 3 x = cos x + sin 2 x + cos 2 x

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

\[ \Leftrightarrow (1 - cos2x) + (sinx - sin2x) + (cos3x - cosx) = 0\]

\[ \Leftrightarrow 2si{n^2}x + (sinx - sin2x) - 2sin2xsinx = 0\]

\[ \Leftrightarrow 2sinx(sinx - sin2x) + (sinx - sin2x) = 0\]

\[ \Leftrightarrow (sinx - sin2x)(2sinx + 1) = 0\]

\( \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{sin2x = sinx}\\{sinx = - \frac{1}{2}}\end{array}} \right.\)

\( \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{2x = x + k2\pi }\\{2x = \pi - x + k2\pi }\\{x = - \frac{\pi }{6} + k2\pi }\\{x = \frac{{7\pi }}{6} + k2\pi }\end{array}} \right. \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = k2\pi }\\{x = \frac{\pi }{3} + \frac{{k2\pi }}{3}}\\{x = - \frac{\pi }{6} + k2\pi }\\{x = \frac{{7\pi }}{6} + k2\pi }\end{array}} \right.\)

Vậy nghiệm của phương trình là:\[x = k2\pi ,x = \frac{\pi }{3} + \frac{{k2\pi }}{3},x = - \frac{\pi }{6} + k2\pi ,x = \frac{{7\pi }}{6} + k2\pi \]