Question

If $3\sin 2\theta =2\sin 3\theta$ and $0<\theta <\pi$, then value of $\sin \theta$ is

• A. $\frac{\sqrt{2}}{3}$
• B. `$\frac{\sqrt{3}}{\sqrt{5}}$
• C. $\frac{\sqrt{15}}{4}$
• D. $\frac{\sqrt{2}}{\sqrt{5}}$

Answer 1

The correct option is C $\frac{\sqrt{15}}{4}$

$3\sin 2\theta =2\sin 3\theta$

$\to 3\times 2\sin \cos \theta =2(3\sin \theta -4\sin 3\theta )$

$\to 6\sin \theta \cos \theta =6\sin \theta -8\sin 3\theta$

$\to \sin \theta (6\cos \theta -6+8{\sin }^2\theta )=0$

$\therefore \sin \theta =0$ or $3\cos \theta +4\sin 2\theta =3$

$\therefore \sin \theta =0$ or $3\cos \theta +4(1-\cos 2\theta )=3$

$\therefore \sin \theta =0$ or $3\cos \theta -4\cos 2\theta +1=0$

$\therefore 4{\cos }^2\theta -3\cos \theta -1=0$

$\therefore 4{\cos }^2\theta -4\cos \theta +\cos \theta -1=0$

$\therefore 4\cos \theta (\cos \theta -1)+1(\cos \theta -1)=0$

$\therefore (4\cos \theta +1)(\cos \theta -1)=0$

$\therefore \cos \theta =\frac{-1}{4}$ or $\cos \theta =1$

$\therefore \sin \theta =\sqrt{1-{\left(\frac{-1}{4}\right)}^2}$

$\therefore \sin \theta =\sqrt{1-\frac{1}{16}}$

$\therefore \sin \theta =\frac{\sqrt{15}}{4}$