Giải phương trình: \[{\rm{cos2x}} - \cos x = \sqrt 3 \left( {\sin 2{\rm{x}} + {\mathop{\rm s}\nolimits} {\rm{inx}}} \right)\].

Lời giải

Ta có:

\[{\rm{cos2x}} - \cos x = \sqrt 3 \left( {\sin 2{\rm{x}} + {\mathop{\rm s}\nolimits} {\rm{inx}}} \right)\]

\( \Leftrightarrow \cos 2x - \sqrt 3 \sin 2x = \sqrt 3 \sin x + \cos x\)

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

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

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

Vậy \[{\rm{x}} = \frac{{ - 2\pi }}{3} + k2\pi ,x = \frac{{k2\pi }}{3},k \in \mathbb{Z}\].