Question

The number of value $x$ in the interval $[0,3\pi ]$ satisfying the $e{q}^{n}2{\sin }^{2}x+5\sin x-3=0$ is

• A. $1$
• B. $2$
• C. $4$
• D. $6$

Answer 1

Answer: (C)

$4$

$2{\sin }^{2}x+5\sin x-3=02{\sin }^{2}x+6\sin x-\sin x-3=02\sin x(\sin x+3)-1(\sin x+3)=0(2\sin x-1)(\sin x+3)=02\sin x-1=0\sin x=\frac{1}{2}x=\frac{\pi }{6},\frac{5\pi }{6},\frac{13\pi }{6},\frac{17\pi }{6}$

So there are four values