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 ...

You visited us 6 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

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}]

No worries! We‘ve got your back. Try BYJU‘S free classes today!

[\frac{π}{3}, \frac{π}{2}]

No worries! We‘ve got your back. Try BYJU‘S free classes today!

[\frac{π}{6}, \frac{3 π}{2}]

No worries! We‘ve got your back. Try BYJU‘S free classes today!

[\frac{π}{6}, \frac{5 π}{6}]

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

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

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

Suggest Corrections

Similar questions

Complete set of values of x satisfying cos2x > |sinx|, $x \in (\frac{- π}{2}, π)$ is

Q. The value of x in

(0, π)

which satisfy the equation

8^{- + | c o s x | + c o s^{2} x + | c o s^{3} x | + \dots t o \infty} = 4^{3}

Q. If

∣ ∣ ∣ 1 - \frac{| sin x |}{1 + | sin x |} ∣ ∣ ∣ \geq \frac{2}{3}

, then

x

lies in

Set of values of $x$ in $(0, π)$ satisfying $1 + {log}_{2} sin x + {log}_{2} sin 3 x \geq 0$ is

Q. The complete set of values of

x

(0, π)

satisfying the equation

1 + {log}_{2} sin x + {log}_{2} sin 3 x \geq 0

Join BYJU'S Learning Program

Explore more

General Solution of Trigonometric Equation

Standard XII Mathematics

Join BYJU'S Learning Program

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

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