The value of x in the interval -0-2 - for which 4 sin 2 x -8 sin x -3 - 0 isA- -6- -2-B- -3- -2-C- -6- 3 -2-D- -6- 5 -6-

The correct option is D [π6, nbsp;5π6]4sin^2 x 8sin x + 3 ≤ 0 ⇒ (2sin x 1)(2sin x 3)≤ 0 As 2sin x lt;3, so for the inequality to be true, sin x≥12 So, f ...

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Standard XII

Mathematics

General Solution of Trigonometric Equation

The value of ...

Question

The value of $x$ in the interval $[0, 2 π]$ for which $4 {sin}^{2} x - 8 sin x + 3 \leq 0$ is

[\frac{π}{6}, \frac{π}{2}]

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[\frac{π}{3}, \frac{π}{2}]

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[\frac{π}{6}, \frac{3 π}{2}]

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[\frac{π}{6}, \frac{5 π}{6}]

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Solution

The correct option is D $[\frac{π}{6}, \frac{5 π}{6}]$
$4 {sin}^{2} x - 8 sin x + 3 \leq 0 \Rightarrow (2 sin x - 1) (2 sin x - 3) \leq 0$
As $2 sin x < 3$ , so for the inequality to be true,
$sin x \geq \frac{1}{2}$
So, from the graph

$∴ x \in [\frac{π}{6}, \frac{5 π}{6}]$

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General Solution of Trigonometric Equation

Standard XII Mathematics

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The value of x in the interval [0,2π] for which 4sin2x−8sinx+3≤0 is

The correct option is D [π6,5π6]4sin2x−8sinx+3≤0⇒(2sinx−1)(2sinx−3)≤0 As 2sinx<3, so for the inequality to be true, sinx≥12 So, from the graph ∴x∈[π6,5π6]

The value of $x$ in the interval $[0, 2 π]$ for which $4 {sin}^{2} x - 8 sin x + 3 \leq 0$ is

The correct option is D $[\frac{π}{6}, \frac{5 π}{6}]$
$4 {sin}^{2} x - 8 sin x + 3 \leq 0 \Rightarrow (2 sin x - 1) (2 sin x - 3) \leq 0$
As $2 sin x < 3$ , so for the inequality to be true,
$sin x \geq \frac{1}{2}$
So, from the graph

$∴ x \in [\frac{π}{6}, \frac{5 π}{6}]$