(A) $\left(0, \frac{\pi}{6}\right) \cup\left(\frac{5\pi}{6}, 2\pi\right)$

(B) $\left(\frac{\pi}{8}, \frac{5\pi}{6}\right)$

(C) $\left(0, \frac{\pi}{8}\right) \cup\left(\frac{\pi}{6}, \frac{5\pi}{6}\right)$

(D) $\left(\frac{41\pi}{48}, \pi\right)$

JEE Advanced · 2006 · 09 Apr 2006 · Trigonometric Equations

0 < θ < 2 π

, then the intervals of values of

θ

for which

2 sin^{2} θ - 5 sin θ + 2 > 0

, is

Answer: (A)

Solution

We need to solve

2 sin^{2} θ - 5 sin θ + 2 > 0

.

Factoring:

(sin θ - 2) (2 sin θ - 1) > 0

.

Since

sin θ \leq 1

, we have

sin θ - 2 < 0

always.

For the product to be positive:

2 sin θ - 1 < 0

, which gives

sin θ < \frac{1}{2}

.

In

(0, 2 π)

sin θ < \frac{1}{2}

when

θ \in (0, \frac{π}{6}) \cup (\frac{5 π}{6}, 2 π)

JEE Advanced · 2006 · 09 Apr 2006 · Single correct option