Question

If $0<\theta <2\pi ,$ then the intervals of values of $\theta$ for which $2{\sin }^2\theta -5\sin \theta +2>0,$ is

• 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. $\left(0,\frac{\pi }{6}\right)\cup \left(\frac{\pi }{6},\frac{5\pi }{6}\right)$
• D. $\left(\frac{41\pi }{48},\pi \right)$

Answer 1

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

R.E.F image

$2si{n}^2\theta -5sin\theta +2>0$

or $2si{n}^2\theta (sin\theta -2)-1(sin\theta -2)>0$

$2sin\theta (sin\theta -2)-1(sin\theta -2)>0$

or $(sin\theta -2)(2sin\theta -1)>0$

or $sin\theta -2$ will be always -ve

ar $|\le sin\theta \le |$

$\therefore (2sin\theta -1)<0$

$sin\theta <\frac{1}{2}$

$\therefore \theta ϵ(0,\pi /6)\cup (\frac{5\pi }{6},2\pi )$

(A) wrong option