A) \[2n\pi \pm \frac{\pi }{4}+\frac{\pi }{12}\]
B) \[n\pi \pm {{(-1)}^{n}}\frac{\pi }{4}+\frac{\pi }{12}\]
C) \[2n\pi \pm \frac{\pi }{4}-\frac{\pi }{12}\]
D) \[n\pi \pm {{(-1)}^{n}}\frac{\pi }{4}-\frac{\pi }{12}\]
Correct Answer: A
Solution :
The given equation is \[(\sqrt{3}-1)\sin \theta +(\sqrt{3}+1)\cos \theta =2\] Let\[(\sqrt{3}-1)=r\sin \alpha \]and\[(\sqrt{3}-1)=r\cos \alpha \] \[\therefore \]\[r=\sqrt{{{(\sqrt{3}-1)}^{2}}+}\overline{{{(\sqrt{3}+1)}^{2}}}=2\sqrt{2}\]and\[\tan \alpha =\frac{\sqrt{3}-1}{\sqrt{3}+1}=\tan ({{60}^{o}}-{{45}^{o}})=\tan {{15}^{o}}\] \[\Rightarrow \] \[\alpha ={{15}^{o}}=\frac{\pi }{12}\] \[\therefore \] \[r\sin \alpha \sin \theta +r\cos \alpha \cos \theta =2\] \[\Rightarrow \] \[2\sqrt{2}\cos (\theta -\alpha )=2\] \[\Rightarrow \] \[\cos (\theta -\alpha )=\frac{1}{\sqrt{2}}=\cos \frac{\pi }{4}\] \[\Rightarrow \] \[\theta \frac{\pi }{12}=2n\pi \pm \frac{\pi }{4}\] \[\Rightarrow \] \[\theta =2n\pi \pm \frac{\pi }{4}+\frac{\pi }{12}\]
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