Giải phương trình \(3\sin x - 4{\sin ^3}x - \sqrt 3 \cos 3x = - 1\).

Lời giải:

\(3\sin x - 4{\sin ^3}x - \sqrt 3 \cos 3x = - 1\)

\( \Leftrightarrow \sin 3x - \sqrt 3 \cos 3x = - 1\)

Chia cả 2 vế của PT trên cho \(\sqrt {{1^2} + {{\left( { - \sqrt 3 } \right)}^2}} = 2\) ta được:

\(\frac{1}{2}\sin 3x - \frac{{\sqrt 3 }}{2}\cos 3x = - \frac{1}{2}\)

\( \Leftrightarrow \sin 3x\cos \frac{\pi }{3} - \sin \frac{\pi }{3}\cos 3x = - \frac{1}{2}\)

\( \Leftrightarrow \sin \left( {3x - \frac{\pi }{3}} \right) = - \frac{1}{2}\)

\( \Leftrightarrow \sin \left( {3x - \frac{\pi }{3}} \right) = \sin \left( { - \frac{\pi }{6}} \right)\)

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

\( \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \frac{\pi }{{18}} + k\frac{{2\pi }}{3}}\\{x = \frac{\pi }{2} + k\frac{{2\pi }}{3}}\end{array}} \right.\left( {k \in \mathbb{Z}} \right)\).