Question

One of the solutions of the equation $8{\sin }^3\theta -7\sin \theta +\sqrt{3}\cos \theta =0$ lies in the interval

• A. $(0^{\circ },{10}^{\circ }]$
• B. $({10}^{\circ },{20}^{\circ }]$
• C. $({20}^{\circ },{30}^{\circ }]$
• D. $({30}^{\circ },{40}^{\circ }]$

Answer 1

The correct option is B $({10}^{\circ },{20}^{\circ }]$

$8{\sin }^3\theta -7\sin \theta +\sqrt{3}\cos \theta =0$
$\Rightarrow 2(3\sin \theta -\sin 3\theta )-7\sin \theta +\sqrt{3}\cos \theta =0$
$\Rightarrow \sqrt{3}\cos \theta -\sin \theta =2\sin 3\theta$
$\Rightarrow \sin \frac{\pi }{3}\cdot \cos \theta -\sin \theta \cdot \cos \frac{\pi }{3}=\sin 3\theta$
$\Rightarrow \sin (\frac{\pi }{3}-\theta )=\sin 3\theta$
$\therefore 4\theta =\frac{\pi }{3}\Rightarrow \theta =\frac{\pi }{12}$
$\theta ={15}^{\circ }$ lies in $({10}^{\circ },{20}^{\circ }]$