Let s = sin x and c = cos x.

LHS = \(\frac{s^3}{c} + \frac{c^3}{s} - 2sc = \frac{s^4 + c^4 - 2s^2c^2}{sc} = \frac{(s^2 - c^2)^2}{sc} = \frac{\cos^2 2x}{sc}\)

RHS = \(1 + \frac{s}{c} + \frac{c}{s} = 1 + \frac{1}{sc}\)

So \(\cos^2 2x = sc + 1\), that is, \(1 - \sin^2 2x = \frac{\sin 2x}{2} + 1\), giving \(\sin 2x\left(\sin 2x + \frac{1}{2}\right) = 0\).

\(\sin 2x = 0\) is not allowed (tan or cot would be undefined). So \(\sin 2x = -\frac{1}{2}\): \(2x = \frac{7\pi}{6}, \frac{11\pi}{6}\).

\(x = \frac{7\pi}{12}, \frac{11\pi}{12}\)

Answer: (C)