Question

Let $S_1$ be the set of all those solutions of the equation $(1+a)\cos \theta \cos (2\theta -b)=(1+a\cos 2\theta )\cos (\theta -b)$ which are independent of $a$ and $b$ and $S_2$ be the set of all such solutions which are dependent on a and $b$.
All the permissible values of $b$, if $a=0$ and $S_2$ is a subset of $(0,\pi )$ is -

• A. $b\in (-n\pi ,2n\pi );n\in Z$
• B. $b\in (-n\pi ,2\pi -n\pi );n\in Z$
• C. $b\in (-n\pi ,n\pi );n\in Z$
• D. none of these

Answer 1

Answer: (B)

$(1+a)\cos \theta \cos (2\theta -b)=(1+a\cos 2\theta )\cos (\theta -b)$
$\Rightarrow \cos \theta \cos (2\theta -b)+a\cos \theta \cos (2\theta -b)$
$=\cos (\theta -b)+a\cos 2\theta \cos (\theta -b)$
$\Rightarrow 2\cos \theta \cos (2\theta -b)+2a\cos \theta \cos (2\theta -b)$
$=2\cos (\theta -b)+2a\cos 2\theta \cos (\theta -b)$
$\Rightarrow \cos (3\theta -b)+\cos (\theta -b)+a(\cos (3\theta -b)+\cos (\theta -b))$
$=2\cos (\theta -b)+a(\cos (3\theta -b)+\cos (\theta +b))$
$\Rightarrow \cos (3\theta -b)+a\cos (\theta -b)=\cos (\theta -b)+a\cos (\theta +b)$
$\Rightarrow \cos (3\theta -b)-\cos (\theta -b)=a(\cos (\theta +b)-\cos (\theta -b))$
$\Rightarrow 2\sin (2\theta -b)\sin \theta =2a\sin \theta \sin b$
$\Rightarrow \sin \theta (\sin (2\theta -b)-a\sin b)=0$
$\Rightarrow \sin \theta =0\text{ or }\sin (2\theta -b)=a\sin b$
$\Rightarrow \sin \theta =0\Rightarrow \theta =n\pi ,n\in Z$
$\text{ and }\sin (2\theta -b)=a\sin b$
$\Rightarrow \sin (2\theta -b)=\sin \left({\sin }^{-1}(a\sin b)\right)$
$\Rightarrow 2\theta -b=n\pi +(-1{)}^n{\sin }^{-1}(a\sin b)n\in Z$
If $a=0$
$\sin (2\theta -b)=0$
$2\theta -b=n\pi n\in Z$
for $S_2$ a subset of $(0,\pi )$
$0<\frac{n\pi +b}{2}<\pi n\in Z$
$\Rightarrow -n\pi <b<2\pi -n\pi$
$b\in (-n\pi ,2\pi -n\pi ),n\in Z$