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Q.Solve the following trigonometrical equation: sin⁡−1x+sin⁡−12x=π3\sin^{-1}x + \sin^{-1}2x = \dfrac{\pi}{3}.

Rajasthan RbseRajasthan Board Senior Secondary Examination 2020Subjective· 3mImportance★★★★★
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Write sin⁡−12x=π3−sin⁡−1x\sin^{-1}2x=\tfrac\pi3-\sin^{-1}x, take sine of both sides using the compound-angle formula, then solve the resulting equation in xx.

Given sin⁡−1x+sin⁡−12x=π3\sin^{-1}x+\sin^{-1}2x=\dfrac\pi3, so sin⁡−12x=π3−sin⁡−1x\sin^{-1}2x=\dfrac\pi3-\sin^{-1}x.

Taking sine of both sides:

2x=sin⁡(π3−sin⁡−1x)=sin⁡π3cos⁡(sin⁡−1x)−cos⁡π3sin⁡(sin⁡−1x)=321−x2−12x2x = \sin\left(\dfrac\pi3-\sin^{-1}x\right) = \sin\dfrac\pi3\cos(\sin^{-1}x) - \cos\dfrac\pi3\sin(\sin^{-1}x) = \dfrac{\sqrt3}{2}\sqrt{1-x^2} - \dfrac12 x

2x+x2=321−x2  ⇒  5x2=321−x2  ⇒  5x=31−x22x+\dfrac{x}{2} = \dfrac{\sqrt3}{2}\sqrt{1-x^2} \;\Rightarrow\; \dfrac{5x}{2}=\dfrac{\sqrt3}2\sqrt{1-x^2} \;\Rightarrow\; 5x=\sqrt3\sqrt{1-x^2}

Squaring both sides: …

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