A) \[2n\pi \pm \frac{\pi }{4}+\frac{\pi }{12}\]
B) \[n\pi +{{(-1)}^{n}}\frac{\pi }{4}+\frac{\pi }{12}\]
C) \[2n\pi \pm \frac{\pi }{4}-\frac{\pi }{12}\]
D) \[n\pi +{{(-1)}^{n}}\frac{\pi }{4}-\frac{\pi }{12}\]
Correct Answer: A
Solution :
Let \[\sqrt{3}+1=r\cos \alpha \] and \[\sqrt{3}-1=r\sin \alpha \]. Then \[r=\sqrt{{{(\sqrt{3}+1)}^{2}}+{{(\sqrt{3}-1)}^{2}}}=2\sqrt{2}\] tan\[\alpha =\frac{\sqrt{3}-1}{\sqrt{3}+1}=\frac{1-(1/\sqrt{3})}{1+(1/\sqrt{3})}=\tan \left( \frac{\pi }{4}-\frac{\pi }{6} \right)\Rightarrow \alpha =\frac{\pi }{12}\] The given equation reduces to \[2\sqrt{2}\cos (\theta -\alpha )=2\Rightarrow \cos \left( \theta -\frac{\pi }{12} \right)=\cos \frac{\pi }{4}\] \[\Rightarrow \] \[\theta -\frac{\pi }{12}=2n\pi \pm \frac{\pi }{4}\Rightarrow \theta =2n\pi \pm \frac{\pi }{4}+\frac{\pi }{12}\].
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