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प्रश्न
Solve the following:
`sin^-1x+sin^-1 2x=pi/3`
उत्तर
We know
`sin^-1x+sin^-1y=sin^-1[xsqrt(1-y^2)+ysqrt(1-x^2)]`
∴ `sin^-1x+sin^-1 2x=pi/3`
⇒ `sin^-1x+sin^-1 2x=sin^-1(sqrt3/2)`
⇒ `sin^-1x-sin^-1(sqrt3/2)=-sin^-1 2x`
⇒ `sin^-1[xsqrt(1-3/4)+sqrt3/2sqrt(1-x^2)]=-sin^-1 2x`
⇒ `sin^-1[x/2+sqrt3/2sqrt(1-x^2)]=sin^-1(-2x)`
⇒ `x/2+sqrt3/2sqrt(1-x^2)=-2x`
⇒ `x+sqrt3sqrt(1-x^2)=-4x`
⇒ `5x=-sqrt3sqrt(1-x^2)`
Squaring both the sides,
`25x^2=3-3x^2`
⇒ `28x^2=3`
⇒ `x=+-1/2sqrt(3/7)`
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