A) \[2n\pi \pm {{(-1)}^{n}}\frac{\pi }{6}\]
B) \[\frac{n\pi }{2}\pm {{(-1)}^{n}}\frac{\pi }{6}\]
C) \[n\pi \pm \frac{\pi }{6}\]
D) \[2n\pi \pm \frac{\pi }{6}\]
Correct Answer: C
Solution :
\[{{\sin }^{2}}\theta =\frac{1}{4}=\]\[{{\sin }^{2}}\frac{\pi }{6}\Rightarrow \theta =n\pi \pm \frac{\pi }{6}\].
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