A) \[n\pi +{{(-1)}^{n}}\frac{\pi }{6}\]
B) \[2n\pi +\frac{\pi }{4}\]
C) \[n\pi +{{(-1)}^{n}}\frac{\pi }{2}\]
D) \[n\pi +{{(-1)}^{n}}\frac{\pi }{3}\]
Correct Answer: B
Solution :
\[{{\sin }^{2}}\theta +\sin \theta -2=0\Rightarrow (\sin \theta -1)\,(\sin \theta +2)=0\] \[\Rightarrow \] \[\sin \theta \ne -2\] , \[\therefore \,\,\sin \theta =1=\sin \pi /2\] \[\Rightarrow \] \[\theta =n\pi +{{(-1)}^{n}}\frac{\pi }{2}\].
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