Giải phương trình: \(\sqrt 3 \cos \left( {x + \frac{\pi }{2}} \right) + \sin \left( {x - \frac{\pi }{2}} \right) = 2\sin 2x\).

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

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

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

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

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

⇔ \[\left[ \begin{array}{l}x + \frac{\pi }{6} = - 2x + k2\pi \\x + \frac{\pi }{6} = \pi + 2x + k2\pi \end{array} \right.\]⇔ \[\left[ \begin{array}{l}x = - \frac{\pi }{{18}} + k\frac{{2\pi }}{3}\\x = - \frac{{5\pi }}{6} - k2\pi \end{array} \right.\left( {k \in \mathbb{Z}} \right)\].