Question

How do you solve `2sin^2(x)+3cos(x)=0` on the interval [0,2pi]?

Answer 1

`(2pi)/3, (4pi/3)`

Explanation:

Replace in the equation `sin^2 x` by `(1 - cos^2 x)`-->
`2(1 - cos^2 x) + 3cos x = 0`
`-2cos^2 x + 3cos x + 2 = 0`
Solve this equation for cos x.
`D = d^2 = b^2 - 4ac = 9 + 16 = 25` --> `d = +- 5`
`cos x = -b/(2a) +- d/(2a) = 3/4 +- 5/-4 = 3/4 +- 5/4`
a. `cos x = 8/4 = 2` (rejected because > 1)
b. `cos x = -2/4 = -1/2`
`cos x = -1/2` --> `x = +- (2pi)/3`
Answers for the interval `(0, 2pi)`:
`(2pi)/3, (4pi)/3` (co-terminal to `(-2pi)/3)`