Question

Solve for x: `sin^-1(x/2) + cos^-1x = π/6`

Answer 1

`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.