A) \[\theta =2n\pi -\frac{\pi }{12}\] or\[\theta =2n\pi +\frac{7\pi }{12}\]
B) \[\theta =n\pi +\frac{\pi }{12}\]
C) \[\theta =2n\pi +\frac{\pi }{12}\] or \[\theta =2n\pi -\frac{7\pi }{12}\]
D) \[\theta =2n\pi +\frac{\pi }{12}\] or \[\theta =2n\pi -\frac{7\pi }{12}\]
Correct Answer: C
Solution :
Given that, \[\cos \theta -sin\theta =\frac{1}{\sqrt{2}}\] \[\Rightarrow \] \[\frac{1}{\sqrt{2}}cos\theta -\frac{1}{\sqrt{2}}\,\sin \theta =\frac{1}{2}\] \[\Rightarrow \] \[\cos \frac{\pi }{4}\cos \theta -\sin \frac{\pi }{4}\sin \theta =\frac{1}{2}\] \[\Rightarrow \] \[\cos \left( \theta +\frac{\pi }{4} \right)=\cos \frac{\pi }{3}\] \[\Rightarrow \] \[\cos \left( \theta +\frac{\pi }{4} \right)=\cos \frac{\pi }{3}\] \[\Rightarrow \] \[\theta +\frac{\pi }{4}=2\pi n\pm \frac{\pi }{3}\] \[\Rightarrow \] \[\theta =2n\pi \pm \frac{\pi }{3}-\frac{\pi }{4}\] \[\Rightarrow \] \[\theta =2n\pi -\frac{7\pi }{12}\] or \[2n\pi +\frac{\pi }{12}\]
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