Question

$\frac{\sin 3\theta -\sin \theta .{\sin }^2\left(2\theta \right)}{\sin \theta +\sin 2\theta .\cos \theta }=\cos x$ . Find the value of x.

• A. $4\theta$
• B. $2\theta$
• C. $\theta$
• D. $3\theta$

Answer 1

Answer: (B)

$2\theta$

$\cos x=\frac{\sin 3\theta -\sin \theta {\sin }^{2}2\theta }{\sin \theta +\sin 2\theta \sin \theta }$
$\Rightarrow \cos x=\frac{(3\sin \theta -4{\sin }^3\theta )-\sin \theta (1-{\cos }^{2}2\theta )}{\sin \theta +\left(2\sin \theta \cos \theta \right)\cos \theta }$
$\Rightarrow \cos x=\frac{3-4{\sin }^2\theta -(1-{\cos }^{2}2\theta )}{2+\cos 2\theta }$
$\Rightarrow \cos x=\frac{3-4\left(\frac{1-\cos 2\theta }{2}\right)-(1-{\cos }^{2}2\theta )}{2+\cos 2\theta }=\frac{2\cos 2\theta +{\cos }^{2}2\theta }{2+\cos 2\theta }$
$\Rightarrow \cos x=\cos 2\theta$
$\therefore x=2\theta$
Hence, option 'B' is correct.