Question

Number of integral solution(s) of the inequality $2{\sin }^{2}x-5\sin x+2>0$ in $x\in [0,2\pi ]$ is-

• A. $3$
• B. $4$
• C. $5$
• D. $6$

Answer 1

The correct option is A $3$

Given

$2si{n}^{2}x-5sinx+2>0$

$\Rightarrow (sinx-2)(2sinx-1)>0$

$2sinx-1<0$

$\Rightarrow sinx<\frac{1}{2}$

$-1\le sinx<\frac{1}{2}$

Hence, The integral solution is $-1,0$

$x=(0,\frac{3\pi }{2},2\pi )$

Hence, The number of integral solution = $3$