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Question 1
Solve: $ 2\sin^{2}\theta - 3\sin\theta - 2 = 0$
A.
$ n\pi + (-1)^{n}\dfrac{\pi}{6} $
B.
$ n\pi + (-1)^{n}\dfrac{\pi}{2} $
C.
$ n\pi + (-1)^{n}\dfrac{5\pi}{6} $
D.
$ n\pi + (-1)^{n}\dfrac{7\pi}{6} $
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