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Question: If \(\sin ^ { - 1 } ( 1 - x ) - 2 \sin ^ { - 1 } x = \pi / 2\) , then x equals....

If sin⁡−1(1−x)−2sin⁡−1x=π/2\sin ^ { - 1 } ( 1 - x ) - 2 \sin ^ { - 1 } x = \pi / 2 , then x equals.

A

(0,−12)\left( 0 , - \frac { 1 } { 2 } \right)

B

(12,0)\left( \frac { 1 } { 2 } , 0 \right)

C

{0}

D

(–1,0)

Answer

{0}

Explanation

Solution

sin⁡−1(1−x)−2sin⁡−1x=π2\sin ^ { - 1 } ( 1 - x ) - 2 \sin ^ { - 1 } x = \frac { \pi } { 2 }

⇒ sin⁡−1(1−x)=(π2−2sin⁡−1x)\sin ^ { - 1 } ( 1 - x ) = \left( \frac { \pi } { 2 } - 2 \sin ^ { - 1 } x \right)

⇒ 1−x=sin⁡(π2−2sin⁡−1x)1 - x = \sin \left( \frac { \pi } { 2 } - 2 \sin ^ { - 1 } x \right)

⇒ 1−x=sin⁡π2cos⁡(2sin⁡−1x)−cos⁡π2sin⁡(2sin⁡−1x)1 - x = \sin \frac { \pi } { 2 } \cos \left( 2 \sin ^ { - 1 } x \right) - \cos \frac { \pi } { 2 } \sin \left( 2 \sin ^ { - 1 } x \right)

⇒ 1−x=cos⁡(2sin⁡−1x)21 - x = \cos \left( 2 \sin ^ { - 1 } x \right) ^ { 2 }

⇒ 1−x=cos⁡cos⁡−1(1−2x1 - x = \cos \cos ^ { - 1 } ( 1 - 2 x ⇒ 2x2−x=02 x ^ { 2 } - x = 0

x=0x = 0, x=1/2x = 1 / 2which does not satisfy the equation

∴ x=0x = 0 is only the solution.