Question

If $4{\sin }^{2}x–8\sin x+3\le 0,$ then the solution set for $x\le 0\le x\le 2\pi$ is

• A. $\left[0,\frac{\pi }{6}\right]$
• B. $\left[0,\frac{5\pi }{6}\right]$
• C. $\left[\frac{5\pi }{6},2\mathrm{\pi }\right]$
• D. $\left[\frac{\pi }{6},\frac{5\pi }{6}\right]$

Answer 1

The correct option is D

$\left[\frac{\pi }{6},\frac{5\pi }{6}\right]$

The explanation for the correct option:

$$\begin{gathered} \Rightarrow 4{\sin }^{2}x–8\sin x+3\le 0 \\ \Rightarrow 4{\sin }^{2}x-2\sin x-6\sin x+3⩽0 \\ \Rightarrow 2\sin x(2\sin x-1)-3(2\sin x-1)⩽0 \\ \Rightarrow (2\sin x-1)(2\sin x-3)⩽0 \end{gathered}$$

Hence,

$\begin{array}{rcl}\sin x& \in & \left[\frac{1}{2},\frac{3}{2}\right]\end{array}$

but we know$\sin x\in [-1,1]$

So required values of $\sin x$ lies in $\begin{array}{rcl}& & \left[\frac{1}{2},1\right]\end{array}$

$\begin{array}{rcl}\therefore x& \in & \left[\frac{\mathrm{\pi }}{6},\frac{5\mathrm{\pi }}{6}\right]\end{array}$

for $0⩽x⩽2\pi$

Hence, the correct answer is option (D).