The solution of the equation sin ^{-1} x+sin | vedclass

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(B) Given equation: $\sin ^{-1} x+\sin ^{-1} 2 x=\frac{\pi}{3}$ Let $x=\sin 	heta$. Then,$	heta+\sin ^{-1}(2 \sin 	heta)=\frac{\pi}{3}$. $\sin ^{-1

Vedclass · Accepted Answer

$\frac{1}{2} \sqrt{\frac{3}{7}}$

Vedclass · Answer

(B) Given equation: $\sin ^{-1} x+\sin ^{-1} 2 x=\frac{\pi}{3}$ Let $x=\sin 	heta$. Then,$	heta+\sin ^{-1}(2 \sin 	heta)=\frac{\pi}{3}$. $\sin ^{-1}(2 \sin 	heta)=\frac{\pi}{3}-	heta$. Taking $\sin$ on both sides: $2 \sin 	heta = \sin \left(\frac{\pi}{3}-	hetaight)$. Using the identity $\sin(A-B) = \sin A \cos B - \cos A \sin B$: $2 \sin 	heta = \sin \frac{\pi}{3} \cos 	heta - \cos \frac{\pi}{3} \sin 	heta$. $2 \sin 	heta = \frac{\sqrt{3}}{2} \cos 	heta - \frac{1}{2} \sin 	heta$. Adding $\frac{1}{2} \sin 	heta$ to both sides: $\frac{5}{2} \sin 	heta = \frac{\sqrt{3}}{2} \cos 	heta$. $	an 	heta = \frac{\sqrt{3}}{5}$. Since $	an 	heta = \frac{	ext{opposite}}{	ext{adjacent}} = \frac{\sqrt{3}}{5}$,the hypotenuse is $\sqrt{(\sqrt{3})^2 + 5^2} = \sqrt{3+25} = \sqrt{28} = 2\sqrt{7}$. Thus,$\sin 	heta = \frac{\sqrt{3}}{2\sqrt{7}} = \frac{1}{2} \sqrt{\frac{3}{7}}$. Since $x = \sin 	heta$,we have $x = \frac{1}{2} \sqrt{\frac{3}{7}}$. Hence,option $B$ is correct.

The solution of the equation $\sin ^{-1} x+\sin ^{-1} 2 x=\frac{\pi}{3}$ is

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