Question

Sol. [Key Concept: $\frac{\sin\ \theta}{6},\cos\ \theta,\ldots$ in G.P. ⇒ ${(middle)}^{2 =}product$; sum the valid $\theta$ in $\lbrack 0,4\pi\rbrack$.]

⇒ $\sin^{2}\theta = 6\cos^{3}\theta \Rightarrow 6\cos^{3}\theta + \cos^{2}\theta - 1 = 0$

$$\begin{matrix} \Rightarrow & \left( 3\cos^{2}\theta + 2cos\theta + 1 \right)(2cos\theta - 1) = 0 \\ \Rightarrow & cos\theta = \frac{1}{2}\theta = \frac{\pi}{3},2\pi - \frac{\pi}{3},2\pi + \frac{\pi}{3},4\pi - \frac{\pi}{3} \\ & Sum = 8\pi \end{matrix}$$