Question

The set of values of $\theta$ satisfying the inequation $2si{n}^2\theta -5sin\theta +2>0$, where 0<$\theta$<2$\pi$,is

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

$2si{n}^2\theta -5sin\theta +2>0$
where $\theta \in (0,2\pi )$
$\Rightarrow (2sin\theta -1)(sin\theta -2)>0$
$sin\theta <\frac{1}{2}andsin\theta >2$
but sin$\theta$ >2
$\therefore sin\theta <\frac{1}{2}$
From the graph of y=sinx, it is obvious that $sinx<\frac{1}{2}$

$ifx\in (0,\frac{\pi }{6})\cup (\frac{5\pi }{6},2\pi ),forx\in (0,2\pi )$

$\therefore sin\theta <\frac{1}{2}\Rightarrow \theta \in (0,\frac{\pi }{6})\cup (\frac{5\pi }{6},2\pi ),for\theta \in (0,2\pi )$