[JEE Advanced 2006] the set of theta in 0 2 pi for which 2 sin 2 theta
The set of $\theta\in(0,2\pi)$ for which $2\sin^2\theta-5\sin\theta+2>0$ is
(a) $\left(0,\frac\pi6\right)\cup\left(\frac{5\pi}6,2\pi\right)$
(b) $\left(\frac\pi8,\frac{5\pi}6\right)$
(c) $\left(0,\frac\pi8\right)\cup\left(\frac\pi6,\frac{5\pi}6\right)$
(d) $\left(\frac{41\pi}{48},\pi\right)$
1 Answer
VAVidaara Admin
✓ Vidaara Team
✓ Accepted
· 2d ago
▲ 0
Correct answer: (a) $\left(0,\frac\pi6\right)\cup\left(\frac{5\pi}6,2\pi\right)$
$2\sin^2\theta-5\sin\theta+2=(2\sin\theta-1)(\sin\theta-2)>0$. Since $\sin\theta-2<0$, need $2\sin\theta-1<0\Rightarrow\sin\theta<\frac12\Rightarrow\theta\in(0,\frac\pi6)\cup(\frac{5\pi}6,2\pi)$.
JEE Advanced 2006 · Trigonometry — verified solution by the Vidaara Team.
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