A) \[x=2n\pi \]
B) \[x=2n\pi +\pi /2\]
C) \[x=n\pi +{{(-1)}^{n}}\frac{\pi }{4}-\frac{\pi }{4}\]
D) \[x=2n\pi \]
E) none of the above
Correct Answer: C
Solution :
We have\[\sin \text{ }x+\cos \text{ }x=1\] \[\Rightarrow \] \[\sqrt{2}\left( \sin x\cos \frac{\pi }{4}+\cos x\sin \frac{\pi }{4} \right)=1\] \[\Rightarrow \] \[\sin \left( x+\frac{\pi }{4} \right)=\frac{1}{\sqrt{2}}\Rightarrow \sin \left( x+\frac{\pi }{4} \right)=\sin \frac{\pi }{4}\] \[\Rightarrow \] \[x+\frac{\pi }{4}=n\pi +{{(-1)}^{n}}\frac{\pi }{4}\] \[\Rightarrow \] \[x=n\pi +{{(-1)}^{n}}\frac{\pi }{4}-\frac{\pi }{4}\]
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