Question

If ${\theta }_1,{\theta }_2,{\theta }_3,{\theta }_4$ are the roots of the equation $\sin (\theta +\alpha )=k\sin 2\theta$, no two of which differ by a multiple of $2\pi$, then ${\theta }_1+{\theta }_2+{\theta }_3+{\theta }_4$ is equal to

• A. $2n\pi ,n\in Z$
• B. $(2n+1)\pi ,n\in Z$
• C. $n\pi ,n\in Z$
• D. None of these

Answer 1

The correct option is B $(2n+1)\pi ,n\in Z$

$\sin (\theta +\alpha )=k\sin 2\theta$
$\sin \theta \cos \alpha +\cos \theta \sin \alpha =2k\sin \theta \cos \theta$
Change the equation in $\tan \frac{\theta }{2}$ form and let $\tan \frac{\theta }{2}=t$, then obtained equation is $\sin \alpha t^4-(2\cos \alpha +4k)t^3+t(4k-2\cos \alpha )-\sin \alpha =0$
$S_1=\frac{2\cos \alpha +4k}{\sin \alpha },S_2=0$
$S_3=\frac{4k-2\cos \alpha }{\sin \alpha },S_4=-1$
$\tan \frac{({\theta }_1+{\theta }_2+{\theta }_3+{\theta }_4)}{2}=\frac{S_1-S_2}{1-S_2+S_4}=\infty$
$\Rightarrow \frac{{\theta }_1+{\theta }_2+{\theta }_3+{\theta }_4}{2}=(2n+1)\frac{\pi }{2}$
$\Rightarrow {\theta }_1+{\theta }_2+{\theta }_3+{\theta }_4=(2n+1)\pi ,n\in$ integer