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Please Solve RD Sharma Class 12 Chapter Inverse Trigonometric Function Exercise 3.12 Question 3 Subquestion (i) Maths Textbook Solution.

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Answer: x= +\frac{1}{2}\sqrt{\frac{3}{7}}
Given: \sin^{-1}x+\sin^{-1}2x= \frac{\pi }{3}
Hint: Using the formula   \sin^{-1}x-\sin^{-1}y= \sin^{-1}\left [ x\sqrt{1-y^{2}}-y\sqrt{1-x^{2}} \right ]
Solution:  We know that  \sin^{-1}\left ( \frac{\sqrt{3}}2{} \right )= \frac{\pi }{3}
\Rightarrow \sin^{-1}2x=\sin^{-1}\frac{\sqrt{3}}{2}-\sin^{-1}x
Using the formula, \sin^{-1}x-\sin^{-1}y= \sin^{-1}\left [ x\sqrt{1-y^{2}}-y\sqrt{1-x^{2}} \right ]
\Rightarrow \sin^{-1}2x= \sin^{-1}\left [ \frac{\sqrt{3}}{2}\sqrt{1-x^{2}}-x\sqrt{1-\left ( \frac{1}{\sqrt{3}} \right )^{2}} \right ]
\Rightarrow 2x=\frac{\sqrt{3}}{2}\sqrt{1-x^{2}}-\frac{x}{2}
\Rightarrow 2x+\frac{x}{2}= \frac{\sqrt{3}}{2}\sqrt{1-x^{2}}
\Rightarrow \frac{5x}{2}= \frac{\sqrt{3}}{2}\sqrt{1-x^{2}}
\Rightarrow 5x= \sqrt{3}\sqrt{1-x^{2}}
Squaring on both sides, we get
\Rightarrow 25x^{2}= 3\left ( 1-x^{2} \right )
\Rightarrow 25x^{2}= 3-3x^{2}
\Rightarrow 28x^{2}= 3x= \pm \frac{1}{2}\sqrt{\frac{3}{7}}
As x= - \frac{1}{2}\sqrt{\frac{3}{7}}is not satisfying the equation
Hence  x= +\frac{1}{2}\sqrt{\frac{3}{7}}

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