[JEE Main 2006] the number of values of x in 0 3 pi satisfying 2 sin 2x
The number of values of $x$ in $[0,3\pi]$ satisfying $2\sin^2x+5\sin x-3=0$ is
(a) $4$
(b) $6$
(c) $1$
(d) $2$
1 Answer
VAVidaara Admin
✓ Vidaara Team
✓ Accepted
· 2d ago
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Correct answer: (a) $4$
$(2\sin x-1)(\sin x+3)=0\Rightarrow\sin x=\frac12$. In $[0,3\pi]$: $\frac\pi6,\frac{5\pi}6,\frac{13\pi}6,\frac{17\pi}6$ — four.
JEE Main 2006 · Trigonometry — verified solution by the Vidaara Team.
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