[JEE Advanced 1995] the general value of theta satisfying 2 sin 2 theta 3 sin theta 2
The general value of $\theta$ satisfying $2\sin^2\theta-3\sin\theta-2=0$ is
(a) $n\pi+(-1)^n\frac\pi6$
(b) $n\pi+(-1)^n\frac\pi2$
(c) $n\pi+(-1)^n\frac{5\pi}6$
(d) $n\pi+(-1)^n\frac{7\pi}6$
1 Answer
VAVidaara Admin
✓ Vidaara Team
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· 2d ago
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Correct answer: (d) $n\pi+(-1)^n\frac{7\pi}6$
$(2\sin\theta+1)(\sin\theta-2)=0\Rightarrow\sin\theta=-\frac12$ (the root $2$ is rejected). The general solution of $\sin\theta=-\frac12=\sin\frac{7\pi}6$ is $n\pi+(-1)^n\frac{7\pi}6$.
JEE Advanced 1995 · Trigonometry — verified solution by the Vidaara Team.
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