If sin^{-1}(1 - x) - 2sin^{-1}x = pi/2,then x | vedclass

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

Question

(C) Given the equation: $\sin^{-1}(1 - x) - 2\sin^{-1}x = \frac{\pi}{2}$ Rearranging the terms,we get: $\sin^{-1}(1 - x) = \frac{\pi}{2} + 2\sin^{-1}x

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(C) Given the equation: $\sin^{-1}(1 - x) - 2\sin^{-1}x = \frac{\pi}{2}$ Rearranging the terms,we get: $\sin^{-1}(1 - x) = \frac{\pi}{2} + 2\sin^{-1}x$ Taking $\sin$ on both sides: $1 - x = \sin\left(\frac{\pi}{2} + 2\sin^{-1}xight)$ Using the identity $\sin(\frac{\pi}{2} + 	heta) = \cos 	heta$: $1 - x = \cos(2\sin^{-1}x)$ Using the identity $\cos(2	heta) = 1 - 2\sin^2	heta$: $1 - x = 1 - 2\sin^2(\sin^{-1}x)$ Simplifying: $1 - x = 1 - 2x^2$ $2x^2 - x = 0$ $x(2x - 1) = 0$ So,$x = 0$ or $x = 1/2$. Checking $x = 1/2$: $\sin^{-1}(1 - 1/2) - 2\sin^{-1}(1/2) = \sin^{-1}(1/2) - 2(\pi/6) = \pi/6 - \pi/3 = -\pi/6 
eq \pi/2$. Checking $x = 0$: $\sin^{-1}(1 - 0) - 2\sin^{-1}(0) = \sin^{-1}(1) - 0 = \pi/2$. Thus,the only solution is $x = 0$.

If $\sin^{-1}(1 - x) - 2\sin^{-1}x = \pi/2$,then $x$ equals:

Similar Questions

Let $f(\theta) = \sin ( \tan ^{-1} ( \frac{\sin \theta}{\sqrt{\cos 2 \theta}} ) )$,where $-\frac{\pi}{4} < \theta < \frac{\pi}{4}$,then the value of $\frac{d}{d(\tan \theta)}(f(\theta))$ is

Given $0 \leq x \leq \frac{1}{2}$,then the value of $\tan \left[\sin ^{-1}\left\{\frac{x}{\sqrt{2}}+\frac{\sqrt{1-x^{2}}}{\sqrt{2}}\right\}-\sin ^{-1} x\right]$ is:

If $y = \sin^2 \left( \cot^{-1} \left( \sqrt{\frac{1-x}{1+x}} \right) \right)$,then $\frac{dy}{dx} = $

$(\tan ^{-1} x)^2+(\cot ^{-1} x)^2=\frac{5 \pi^2}{8} \Rightarrow x=$

The value of $\tan ^{-1}\left\{\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right\}+\frac{1}{2} \cos ^{-1} x$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module