Question

How do you solve `sin^2x + 2sinx - 3 = 0`?

Answer 1

Let sin x = t,

`f(t) = t^2 + 2t - 3 = 0`

Since a + b + c = 0, one real root is t = 1 and the other: `t = c/a = -3.`

`Solve: t = sin x = 1 = sin (pi/2) -> x = pi/2`

The solution t = -3 is rejected.