(b)

Given,

$8 \sin ^3 \theta-7 \sin \theta+\sqrt{3} \cos \theta =0$

$2\left(4 \sin ^3 \theta\right)-7 \sin \theta+\sqrt{3} \cos \theta =0$

$2(3 \sin \theta-\sin 3 \theta)-7 \sin \theta$

$+\sqrt{3} \cos \theta =0$

$6 \sin \theta-2 \sin 3 \theta-7 \sin \theta$

$+\sqrt{3} \cos \theta =0$

$\sqrt{3} \cos \theta-\sin \theta-2 \sin 3 \theta =0$

$\frac{\sqrt{3}}{2} \cos \theta-\frac{1}{2} \sin \theta =\sin 3 \theta$

$\sin \left(\frac{\pi}{3}-\theta\right) =\sin 3 \theta$

$\frac{\pi}{3}-\theta=3 \theta \Rightarrow 4 \theta=60^{\circ} \Rightarrow \theta =15^{\circ}$

$\theta \in\left(10^{\circ}, 20^{\circ}\right]$