Question

How do you find all solutions of the equation in the interval `[0,2pi)` given `2sin^2x-3sinx=-1`?

Answer 1

`pi/6; pi/2, (5pi)/6`

Explanation:

Solve this quadratic equation for sin x:
`2sin^2 x - 3sin x + 1 = 0`
Since a + b + c = 0, use shortcut. the 2 real roots are:
`sin x = 1` and `sin x = c/a = 1/2`
a. sin x = 1 , unit circle gives --> `x = pi/2`
b. `sin x = 1/2` , trig table and unit circle give -->
`x = pi/6` and `x = pi - pi/6 = (5pi)/6`
Answers for `(0, 2pi)`:
`pi/6; pi/2, (5pi)/6`