Question

Solve the equation `sin^-1 6x + sin^-1 6sqrt(3)x = - pi/2`

Answer 1

From the given equation

we have `sin^-1 6x = - pi/2 - sin^-1 6sqrt(3)x`

⇒ `sin(sin^-1 6x) = sin(- pi/2 - sin^-1 6sqrt(3)x)`

⇒ 6x = `- cos(sin^-1 6sqrt(3)x)`

⇒ 6x = `-sqrt(1 - 108x^2)`.

Squaring, we get

`36x^2= 1 - 108x^2`

⇒ 144x² = 1

⇒ x = `+- 1/12`

Note that x = `- 1/12` is the only root of the equation as x = `1/12` does not satisfy it.