Question

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

Answer 1

`{pi/2}`

Explanation:

Factor to solve.

`(sinx - 1)(sinx - 2) = 0`

`sinx = 1 and sinx = 2`

`x = pi/2 and x = O/`

Hence, the solution set is `{pi/2}`.

Hopefully this helps!