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प्रश्न
Solve for x: `sin^-1(x/2) + cos^-1x = π/6`
उत्तर
`sin^-1(x/2) + cos^-1x = π/6`
Since, `sin^-1x + cos^-1x = π/2`
∴ `sin^-1(x/2) + π/2 - sin^-1x = π/6`
`\implies -sin^-1x + sin^-1 x/2 = π/6 - π/2`
`\implies -sin^-1x + sin^-1 x/2 = (2π - 6π)/12`
`\implies -sin^-1x + sin^-1 x/2 = -π/3`
`\implies sin^-1 x/2 = -π/3 + sin^-1x`
`\implies x/2 = sin(-π/3 + sin^-1x)`
`\implies x/2 = sin((-π)/3)cos(sin^-1x) + cos((-π)/3)sin(sin^-1x)`
`\implies x/2 = -sin π/3 cos cos^-1 sqrt(1 - x^2) + cos(π/3)x`
`\implies x/2 = -sqrt(3)/2 sqrt(1 - x^2) + x/2`
`\implies 0 = - sqrt(3)/2 sqrt(1 - x^2)`
`\implies` 1 – x2 = 0
`\implies` x2 = 1
∴ x = 1 is the only answer because x = – 1 will not satisfy above question.
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