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. $b \in(- n \pi, 2 n \pi) ; n \in Z$
• 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)

$(1+a) \cos \theta \cos (2 \theta-b)=(1+a \cos 2 \theta) \cos (\theta-b)$
$\Rightarrow \cos \theta \cos (2 \theta-b)+a \cos \theta \cos (2 \theta-b)$
$=\cos (\theta-b)+a \cos 2 \theta \cos (\theta-b)$
$\Rightarrow 2 \cos \theta \cos (2 \theta-b)+2 a \cos \theta \cos (2 \theta-b)$
$=2 \cos (\theta- b )+2 a \cos 2 \theta \cos (\theta- b )$
$\Rightarrow \cos (3 \theta-b)+\cos (\theta-b)+a(\cos (3 \theta-b)+\cos (\theta-b))$
$=2 \cos (\theta-b)+a(\cos (3 \theta-b)+\cos (\theta+b))$
$\Rightarrow \cos (3 \theta-b)+a \cos (\theta-b)=\cos (\theta-b)+a \cos (\theta+b)$
$\Rightarrow \cos (3 \theta-b)-\cos (\theta-b)=a(\cos (\theta+b)-\cos (\theta-b))$
$\Rightarrow 2 \sin (2 \theta-b) \sin \theta=2 a \sin \theta \sin b$
$\Rightarrow \sin \theta(\sin (2 \theta-b)-a \sin b)=0$
$\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$