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\[\cos 2\theta=1-2 \sin ^2\] \[\cos 2\theta=\frac{ \sqrt{3} }{ 2}=\cos \frac{ \pi }{ 6 },\cos (2\pi-\frac{ \pi }{ 6})\] \[\cos 2\theta=\cos (2n \pi+\frac{ \pi }{ 6}),\cos (2n \pi+\frac{ 11\pi }{ 6})\] \[2\theta=2n \pi+\frac{ \pi }{ 6 },2n \pi+\frac{ 11\pi }{ 6}\] \[\theta=n \pi+\frac{ \pi }{ 12 },n \pi+\frac{ 11\pi }{ 12 },n \in ~R\]
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so it would be option D.) ?
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