Tìm x, biết: \(\sqrt 3 \cos \left( {x - \frac{\pi }{2}} \right) + \sin \left( {x - \frac{\pi }{2}} \right) = 2\sin 2x\).

\(\sqrt 3 \cos \left( {x - \frac{\pi }{2}} \right) + \sin \left( {x - \frac{\pi }{2}} \right) = 2\sin 2x\)

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

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

\( \Leftrightarrow \sin \left( {x - \frac{\pi }{6}} \right) = \sin 2x\)

\( \Leftrightarrow \left[ \begin{array}{l}x - \frac{\pi }{6} = 2x + k2\pi \\x - \frac{\pi }{6} = \pi - 2x + k2\pi \end{array} \right.\)(k ∈ ℤ)

\( \Leftrightarrow \left[ \begin{array}{l}x = - \frac{\pi }{6} + k2\pi \\x = \frac{{7\pi }}{{18}} + \frac{{k2\pi }}{3}\end{array} \right.\)(k ∈ ℤ)

Vậy phương trình đã cho có hai họ nghiệm là: \(\left[ \begin{array}{l}x = - \frac{\pi }{6} + k2\pi \\x = \frac{{7\pi }}{{18}} + \frac{{k2\pi }}{3}\end{array} \right.\).