If x∈[0,2π], then the solution set of the inequation 4sin2⁡x−8sin⁡x+3≤0, is

A
$[0, π / 6]$
B
$[0, 5 π / 6]$
C
$[5 π / 6, 2 π]$
D
$[π / 6, 5 π / 6]$

Solution:

we have ,

$\begin{array}{l} 4 \sin^{2} x - 8 \sin x + 3 \leq 0 \\ \Rightarrow (2 \sin x - 1) (2 \sin x - 3) \leq 0 \\ \Rightarrow 2 \sin x - 1 \geq 0 [∵ 2 \sin x - 3 \leq 0 for all x] \end{array}$

$\Rightarrow \sin x \geq \frac{1}{2}$

$\Rightarrow x \in [π / 6, 5 π / 6]$ $[D r a w g r a p h o f y = \sin x a n d f i n d t h e v a l u e s]$

If $x \in [0, 2 π],$ then the solution set of the inequation $4 \sin^{2} x - 8 \sin x + 3 \leq 0$ , is

Solution:

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