Giải phương trình: \(8\sin x = \frac{{\sqrt 3 }}{{\cos x}} + \frac{1}{{\sin x}}\).

\(8\sin x = \frac{{\sqrt 3 }}{{\cos x}} + \frac{1}{{\sin x}}\)

⇔ \(8\sin x = \frac{{\sqrt 3 \sin x + \cos x}}{{\cos x.\sin x}}\)

Điều kiện: cosxsinx ≠ 0 hay sin2x ≠ 0

Suy ra: 4sinxsin2x = \(\sqrt 3 \sin x + \cos x\)

⇔ 2(cosx – cos3x) = \(\sqrt 3 \sin x + \cos x\)

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

⇔ \(\cos \left( {\frac{\pi }{3}} \right)\cos x - \sin \left( {\frac{\pi }{3}} \right)\sin x = \cos 3x\)

⇔ \(\cos \left( {\frac{\pi }{3} + 3x} \right) = \cos 3x\)

⇔ \[\left[ \begin{array}{l}3x = \frac{\pi }{3} + 3x + k2\pi \Leftrightarrow x = \frac{\pi }{6} + k\pi \\3x = - x - \frac{\pi }{3} + k2\pi \Leftrightarrow x = - \frac{\pi }{{12}} + k\frac{\pi }{2}\end{array} \right.\left( {k \in \mathbb{Z}} \right)\].