A) \[\{n\pi ,n\in Z\}\]and\[\frac{1}{2}\]\[\{n\pi +{{(-1)}^{n}}si{{n}^{-1}}(asinb)+b;n\in Z\}\]
B) \[\{n\frac{\pi }{2},n\in Z)\]and\[\{n\pi +{{(-1)}^{n}}si{{n}^{-1}}(asinb);n\in Z\}\]
C) \[\{n\frac{\pi }{2},n\in Z\}\]and\[\{n\pi +{{(-1)}^{n}}si{{n}^{-1}}(\frac{a}{2}sinb);n\in Z\}\]
D) None of these
Correct Answer: A
Solution :
\[(1+a)cos\theta \cos (2\theta -b)=(1+acos2\theta )cos(\theta -b)\]\[\Rightarrow \]\[cos\theta \cos (2\theta -b)+a\,cos\theta cos(2\theta -b)\] \[=cos(\theta -b)+a\,cos2\theta cos(\theta -b)\] \[\Rightarrow \cos (3\theta -b)+cos(\theta -b)+\]\[a[cos(3\theta -b)+cos(\theta -b)]\] \[=2\cos (\theta -b)+a\{\cos (3\theta -b)+cos(\theta -b)]\}\] \[\Rightarrow \]\[\cos (3\theta -b)-\cos (\theta -b)=a\,cos(\theta -b)-acos(\theta -b)\]\[\Rightarrow \]\[2sin(\theta -b)sin\theta =2a\sin \theta \sin b\] \[\Rightarrow \]\[sin\theta =0or\sin (2\theta -b)=asinb\] \[\sin \theta =0\Rightarrow \theta =n\pi ,n\in Z;\]\[\Rightarrow \]\[{{S}_{1}}=\{n\pi |n\in Z\}\] and\[2\theta -b=n\pi +{{(-1)}^{n}}{{\sin }^{-1}}(asinb),n\in Z\] \[\Rightarrow \]\[{{S}_{2}}=\frac{1}{2}\{n\pi +{{(-1)}^{n}}si{{n}^{-1}}(asinb)+b,n\in Z\}\]
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