A) \[2n\pi \]
B) \[n\pi +{{(-1)}^{n}}\frac{\pi }{4}-\frac{\pi }{4}\]
C) \[2n\pi +\frac{\pi }{2}\]
D) none of these
Correct Answer: B
Solution :
Since, \[\sin \theta +\cos \theta =1\] \[\Rightarrow \]\[\sqrt{2}\left( \frac{1}{\sqrt{2}}\sin \theta +\frac{1}{\sqrt{2}}\cos \theta \right)=1\] \[\Rightarrow \] \[\sqrt{2}\sin \left( \frac{\pi }{4}+\theta \right)=1\] \[\Rightarrow \] \[\sin \left( \frac{\pi }{4}+\theta \right)=\frac{1}{\sqrt{2}}=\sin \left( \frac{\pi }{4} \right)\] \[\therefore \] The general value is \[\frac{\pi }{4}+\theta =n\pi +{{(-1)}^{n}}\frac{\pi }{4}\] \[\Rightarrow \] \[\theta =n\pi +{{(-1)}^{n}}\frac{\pi }{4}-\frac{\pi }{4}\]
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