A) \[x=2n\pi -\frac{\pi }{12}\] or \[x=2n\pi +\frac{7\pi }{12}\]
B) \[x=n\pi \pm \frac{\pi }{12}\]
C) \[x=2n\pi +\frac{\pi }{12}\] or \[x=2n\pi -\frac{7\pi }{12}\]
D) \[x=n\pi +\frac{\pi }{12}\] or \[x=n\pi -\frac{7\pi }{12}\]
Correct Answer: C
Solution :
Given equation is,\[\cos x-\sin x=\frac{1}{\sqrt{2}}\] Dividing equation by \[\sqrt{2}\], \[\frac{1}{\sqrt{2}}\cos x-\frac{1}{\sqrt{2}}\sin x=\frac{1}{2}\] \[\cos \left( \frac{\pi }{4}+x \right)=\cos \frac{\pi }{3}\]. Hence, \[\frac{\pi }{4}+x=2n\pi \pm \frac{\pi }{3}\] \[x=2n\pi +\frac{\pi }{3}-\frac{\pi }{4}=2n\pi +\frac{\pi }{12}\] or\[x=2n\pi -\frac{\pi }{3}-\frac{\pi }{4}=2n\pi -\frac{7\pi }{12}\].
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