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Question
If `sin^-1(1-x) -2sin^-1x = pi/2` then x is
- -1/2
- 1
- 0
- 1/2
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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