Question

How do you solve `2sin^2(x)+cos(x)=1` over the interval 0 to 2pi?

Answer 1

`0, (2pi)/3, (4pi)/3, and 2pi`

Explanation:

Replace in the equation `sin^2 x` by `(1 - cos^2 x)`-->
`2 - 2cos^2 x + cos x = 1`
`2cos^2 x - cos x - 1 = 0`.
Solve this equation for cos x.
Since a + b + c = 0, use shortcut. The 2 real roots are; cos x = 1 and `cos x = c/a = -1/2`
a. `cos x = 1` --> x = 0, and `x = 2pi`
b. `cos x = -1/2` --> `x = pi/3` and `x = -pi/3 (or (4pi)/3)`
Answer for interval `(0, 2pi)`:
`0, (2pi)/3, (4pi)/3, 2pi.`
Check.
If x = (2pi)/3 --> 2sin^2 x = 3/2 --> cos x = -1.2 --> 3/2 - 1/2 = 1. OK
If x = (4pi)/3 --> 2sin^2 x = 3/2 --> cos x = -1/2 --> 3/2 - 1/2 = 1. OK