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प्रश्न
Solve the following:
`cos^-1x+sin^-1 x/2=π/6`
उत्तर
`cos^-1x+sin^-1 (x/2) = π/6`
`(π/2 − sin^-1x) + sin^-1(x/2) = π/6`
`π/2 − π/6 = sin^-1x−sin^-1(x/2)`
`sin^-1x−sin^-1(x/2) = π/3 = sin^-1(sqrt3/2)`
`sin^-1x = sin^-1(sqrt3/2) + sin^-1(x/2)`
`sin^-1(x) = sin^-1(sqrt3/2 sqrt(1−x^2/4) + x/2 sqrt(1 - 3/4)) ...[sin^-1 x + sin^-1y = sin^-1 [xsqrt(1 - y^2) + ysqrt(1 - x^2)].`
`sin^-1(x) = sin^-1[(sqrt3/2 sqrt(4−x^2)/2) + x/2 . 1/2]`
`x = (sqrt3 sqrt(4 - x^2))/4 + x/4`
`x - x/4 = (sqrt3 sqrt(4 - x^2))/4`
`(3x)/cancel4 = (sqrt3 sqrt(4 - x^2))/cancel4`
Squaring both the sides
9x2 = 3(4 − x2)
3x2 = 4 - x2
3x2 + x2 = 4
4x2 = 4
x2 = 1
x = ± 1.
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