A) \[b\in (-n\pi ,2n\pi ),n\in Z\]
B) \[b\in (-n\pi ,2-n\pi ),n\in Z\]
C) \[b\in (-n\pi ,n\pi ),n\in Z\]
D) none of these
Correct Answer: B
Solution :
\[a=0\Rightarrow \sin (2\theta -b)=0\] \[2\theta -b=n\pi ,n\in Z\] \[\theta =\frac{n\pi +b}{2},n\in Z:{{S}_{2}}=\left[ \frac{n\pi +b}{2};n\in Z \right]\] \[{{S}_{2}}\] must be subject of \[(0,\pi )\] \[0<\frac{n\pi +b}{2}<\pi ,n\in Z\] \[0<n\pi +b<2\pi \Rightarrow -n\pi <b<-n\pi +2\pi \] \[\Rightarrow \]\[b\in (-n\pi ,2\pi -n\pi ),n\in Z.\]You need to login to perform this action.
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