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

Answer» `(pi/2-sin^(-1)x)+sin^(-1)(x/2)=pi/6`
`pi/22-pi/6=sin^(-1)x-sin^(-1)(x/2)`
`sin^(-1)x-sin^(-1)(x/2)=pi/3=sin^(-1)(sqrt3/2)`
`sin^(-1)x=sin^(-1)(sqrt3/2)+sin^(-1)(x/2)`
`sin^(-1)(x)=sin^(-1)(sqrt3/2sqrt(1-x^2/4)+x/2sqrt(1-3/4))`
`sin^(-1)x=sin^(-1)(sqrt3/2sqrt(4-x^2)/2+x/2*1/2]`
`x=(sqrt3sqrt(4-x^2))/4+x/4`
`sqrt3/4x=sqrt3/4sqrt(4-x^2`
`3x^2=4-x^2`
`x^2=1`
`x=pm1`.


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