Question

How do you solve the following equation `-2sin^2x=sin x-1` in the interval [0, 2pi]?

Answer 1

`pi/6; (5pi)/6; (3pi)/2`

Explanation:

Bring the quadratic equation to standard form, then solve it for sin x:
`2sin^2 x + sin x - 1 = 0`
Since a - b + c = 0, use shortcut. One real root is sin x = -1 and the other is `sin x = -c/a = 1/2`
Use trig table and unit circle:
a. sin x = -1 --> arc `x = (3pi)/2`
b. `sin x = 1/2` -->
arc `x = pi/6` and arc `x = (5pi)/6`

Answers for `(0, 2pi)`
`pi/6; (5pi)/6, (3pi)/2`