Question

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

• A. $6$
• B. $1$
• C. $2$
• D. $4$

Answer 1

The correct option is D

$4$

Explanation for the correct answer:

Given, $2{\sin }^{2}x+5\sin x-3=0$

$$\begin{gathered} \Rightarrow 2{\sin }^{2}x+6\sin x–\sin x–3=0 \\ \Rightarrow 2\sin \left(x\right)\left(\sin x+3\right)-1(\sin x+3)=0 \\ \Rightarrow (2\sin x-1)\left(\sin x+3\right)=0 \end{gathered}$$

$\sin x=\frac{1}{2},\sin x=-3$ {not possible since $-1\le \sin x\le 1$}

Thus, $x=\frac{\mathrm{\pi }}{6},\frac{5\mathrm{\pi }}{6},\frac{13\mathrm{\pi }}{6},\frac{17\mathrm{\pi }}{6}$

Hence, the number of values of $x$ in the interval $[0,3\pi ]$are $4.$

Hence, option D is the correct answer.