Question

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

Answer 1

`pi/6 and (5pi)/6`

Explanation:

`f(x) = 2sin^2 x + 3sin x - 2 = 0`
Replace (3sin x) by (4sin x - sin x)
f(x) = 2sin^2 x + 4sin x - sin x - 2 = 0
Factor by grouping:
f(x) = 2sin x (sin x + 2) - (sin x + 2) = 0
f(x) = (sin x + 2)(2sin x - 1) = 0
There are 2 solutions:
a. sin x = - 2 (rejected since < -1)
b. `sin x = 1/2`
The trig table and unit circle give 2 arcs x:
`x = pi/6` and `x = (5pi)/6` that have same sine value `(1/2)`
Answers for`(0, 2pi)`:
`pi/6 and (5pi)/6`