#### Question

If `sin^-1(1-x) -2sin^-1x = pi/2` then x is

- -1/2
- 1
- 0
- 1/2

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#### Solution

**(c)**

`sin^-1(1-x)-2sin^-1x=pi/2`

`sin^-1(1-x)=pi/2+2sin^-1x`

`(1-x)=sin(pi/2+2sin^-1x)`

`(1-x)=cos(2sin^-1x)`

`(1-x)=cos(cos^-1(1-2x^2))`

`(1-x)=1-2x^2`

`2x^2-x=0`

`x(2x-1)=0`

`x=0 or 2x-1=0`

`x=0 or x=1/2`

`"for "x =1/2`

`sin^-1(1-x)-2sin^-1x=sin^-1(1/2)-2sin^-1(1/2)=-sin^-1(1/2)=pi/6`

So x=1/2 is not solution of the given equation

for x=0

`sin^-1(1-x)-2sin^-1x=sin^-1(1)-2sin^-1(0)=pi/2-0=pi/2`

So x = 0 is a valid solution of the given equation.

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#### Reference Material

Solution for question: If sin^−1(1−x)−2sin^−1 x=π/2 then x is concept: Basic Concepts of Trigonometric Functions. For the courses HSC Science (Computer Science), HSC Science (General) , HSC Science (Electronics), HSC Arts