Question

How do you solve `2sin^2 x=2+cos x` between 0 and 2pi?

Answer 1

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

Explanation:

Replace in the equation `sin^2 x` by `(1 - cos^2 x`
`2(1 - cos^2 x) = 2 + cos x`
`- 2cos^2 x = cos x`
`cos x + 2cos^2 x = cos x(1 + 2cos x) = 0`
a. cos x = 0 --> `x = pi/2 and x = (3pi)/2`
b. (1 + 2cos x) = 0 --> `cos x = - 1/2` -->`x = +- (2pi)/3`
The co-terminal arc of - (2pi)/3 is arc (4pi)/3.
Answer for `(0, 2pi)`:
`pi/2, (2pi)/3, (4pi)/3, (3pi)/2`