Number of real values of x in (0, pi) for whi | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given the inequality: $\frac{8}{3\sin x - \sin 3x} + 3\sin^2 x \le 5$.Using the identity $\sin 3x = 3\sin x - 4\sin^3 x$,the denominator becomes $

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(B) Given the inequality: $\frac{8}{3\sin x - \sin 3x} + 3\sin^2 x \le 5$.Using the identity $\sin 3x = 3\sin x - 4\sin^3 x$,the denominator becomes $3\sin x - (3\sin x - 4\sin^3 x) = 4\sin^3 x$.Substituting this into the inequality: $\frac{8}{4\sin^3 x} + 3\sin^2 x \le 5 \Rightarrow \frac{2}{\sin^3 x} + 3\sin^2 x \le 5$.Let $f(x) = \frac{2}{\sin^3 x} + 3\sin^2 x - 5$. We want to find $x \in (0, \pi)$ such that $f(x) \le 0$.Let $t = \sin x$. Since $x \in (0, \pi)$,$t \in (0, 1]$.The expression becomes $g(t) = \frac{2}{t^3} + 3t^2 - 5$.By the $AM$-$GM$ inequality for three terms $\frac{1}{t^3}, \frac{1}{t^3}, t^2, t^2, t^2$:$\frac{\frac{1}{t^3} + \frac{1}{t^3} + t^2 + t^2 + t^2}{5} \ge \sqrt[5]{\frac{1}{t^3} \cdot \frac{1}{t^3} \cdot t^2 \cdot t^2 \cdot t^2} = \sqrt[5]{1} = 1$.So,$\frac{2}{t^3} + 3t^2 \ge 5$.The equality holds if and only if $\frac{1}{t^3} = t^2$,which implies $t^5 = 1$,so $t = 1$.Since $\sin x = 1$ for $x \in (0, \pi)$ only at $x = \frac{\pi}{2}$,the inequality $f(x) \le 0$ holds only for $x = \frac{\pi}{2}$.Thus,there is only $1$ real value.

Number of real values of $x \in (0, \pi)$ for which $\frac{8}{3\sin x - \sin 3x} + 3\sin^2 x \le 5$ is:

Similar Questions

The integer part of the number $\sum_{k=0}^{44} \frac{1}{\cos k^{\circ} \cos (k+1)^{\circ}}$ is

If $\frac{\cos^4 \alpha}{\cos^2 \beta} + \frac{\sin^4 \alpha}{\sin^2 \beta} = 1$,then the value of $\left[ \frac{\cos^4 \beta}{\cos^2 \alpha} + \frac{\sin^4 \beta}{\sin^2 \alpha} \right]$ is (where $[.]$ denotes the greatest integer function).

The equation $\frac{\sin^3 \theta - \cos^3 \theta}{\sin \theta - \cos \theta} - \frac{\cos \theta}{\sqrt{1 + \cot^2 \theta}} - 2 \tan \theta \cot \theta = -1$ holds true if:

If $\sinh x = \frac{12}{5}$,then $\sinh 3x + \cosh 3x = $

Let $x=a \sin ^\alpha \theta \cos ^{\alpha+1} \theta$ and $y=a \sin ^{\alpha+1} \theta \cos ^\alpha \theta$,where $\theta \neq \frac{n \pi}{2}$. If $\frac{(x^2+y^2)^m}{(xy)^n}$ is independent of $\theta$,then the relation between $\alpha, m$ and $n$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module