Question

How do you solve `2sin^2x + sinx =0` from `[0,2pi]`?

Answer 1

Solve `2sin^2 x + sin x = 0` `[0, 2pi]`

Explanation:

`sin x(2sin x + 1) = 0`
Use trig table and unit circle -->

a. sin x = 0 => x = 0, and `x = pi` and `x = 2pi`

b. 2sin x + 1 = 0 => `sin x = -1/2` => `x = (7pi)/6` and `x = (11pi)/6`.

Finally, within interval `[0, 2pi],` there are 5 answers:
`0, pi, 2pi, (7pi)/6 and (11pi)/6`