A) \[2n\pi \pm \frac{\pi }{6}\]
B) \[2n\pi \pm \frac{\pi }{4}\]
C) \[n\pi +{{(-1)}^{n}}\frac{\pi }{3}\]
D) \[n\pi +{{(-1)}^{n}}\frac{\pi }{4}\]
Correct Answer: C
Solution :
We have,\[2\sqrt{3}\cos \theta =\tan \theta \] \[\Rightarrow \] \[2\sqrt{3}\cos \theta =\frac{\sin \theta }{\cos \theta }\] \[\Rightarrow \] \[2\sqrt{3}{{\cos }^{2}}\theta =\sin \theta \] \[\Rightarrow \] \[2\sqrt{3}{{\sin }^{2}}\theta +\sin \theta -2\sqrt{3}=0\] \[\Rightarrow \] \[\sin \theta =\frac{-1\pm 7}{4\sqrt{3}}\] \[\Rightarrow \] \[\sin \theta =\frac{-8}{4\sqrt{3}}\](impossible) \[\therefore \] \[\sin \theta =\frac{6}{4\sqrt{3}}=\frac{\sqrt{3}}{2}\] \[\Rightarrow \] \[\theta =n\pi +{{(-1)}^{n}}\frac{\pi }{3}\]
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