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Mathematics Question on Properties of Inverse Trigonometric Functions

The solution of sin⁡−1 x−sin⁡−1 2x=±π3\sin^{-1}\,x-\sin^{-1}\,2x=\pm\frac{\pi}{3} is

A

±13\pm\frac{1}{3}

B

±14\pm\frac{1}{4}

C

±32\pm\frac{\sqrt{3}}{2}

D

±12\pm\frac{1}{2}

Answer

±12\pm\frac{1}{2}

Explanation

Solution

sin⁡−1x−sin⁡−12x=±π3\sin ^{-1} x-\sin ^{-1} 2 x=\pm \frac{\pi}{3}
⇒sin⁡−1x−sin⁡−1(±32)=sin⁡−12x\Rightarrow \sin ^{-1} x-\sin ^{-1}\left(\pm \frac{\sqrt{3}}{2}\right)=\sin ^{-1} 2 x
⇒sin⁡−1[x1−34−(±321−x2)]\Rightarrow \sin ^{-1}\left[x \sqrt{1-\frac{3}{4}}-\left(\pm \frac{\sqrt{3}}{2} \sqrt{1-x^{2}}\right)\right]
=sin⁡−12x=\sin ^{-1} 2 x
⇒x2−(±321−x2)=2x\Rightarrow \frac{x}{2}-\left(\pm \frac{\sqrt{3}}{2} \sqrt{1-x^{2}}\right)=2 x
⇒−(±31−x2)=3x\Rightarrow-\left(\pm \sqrt{3} \sqrt{1-x^{2}}\right)=3 x
On squaring, both sides we get,
3(1−x2)=9x23\left(1-x^{2}\right)=9 x^{2}
⇒1−x2=3x2\Rightarrow 1-x^{2}=3 x^{2}
⇒4x2=1\Rightarrow 4 x^{2}=1
⇒x=±12\Rightarrow x=\pm \frac{1}{2}