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Question: The solution set of the equation \(\sin^{- 1}x = 2\tan^{- 1}x\)is...

The solution set of the equation sin⁡−1x=2tan⁡−1x\sin^{- 1}x = 2\tan^{- 1}xis

A

{1, 2}

B

{−1,2}\{ - 1,2\}

C

{−1,1,0}\{ - 1,1,0\}

D

{1,1⥂/⥂2,0}\{ 1,1 ⥂ / ⥂ 2,0\}

Answer

{−1,1,0}\{ - 1,1,0\}

Explanation

Solution

sin⁡−1x=2tan⁡−1x\sin^{- 1}x = 2\tan^{- 1}x ⇒ sin⁡−1x=sin⁡−12x1+x2\sin^{- 1}x = \sin^{- 1}\frac{2x}{1 + x^{2}}

⇒ 2x1+x2=x\frac{2x}{1 + x^{2}} = x

⇒ x3−x=0x^{3} - x = 0 ⇒ x(x+1)(x−1)=0⇒x={−1,1,0}x(x + 1)(x - 1) = 0 \Rightarrow x = \{ - 1,1,0\}