Question

How do you solve `2sin^2a=2+cosa` and find all solutions in the interval `[0,2pi)`?

Answer 1

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

Explanation:

Replace in the equation `2sin^2 a` by `2(1 - cos^2 a)`, we get:
`2 - 2cos^2 a = 2 + cos a`
`-2cos^2 a - cos a = 0`
`- cos a(2cos a + 1) = 0`
Trig table and unit circle -->
a. `cos a = 0` --> `a = pi/2` and `a = (3pi)/2`
b. 2cos a + 1 = 0
`cos a = -1/2` --> `a = +- (2pi)/3`
The arc`(-2pi)/3` and arc `(4pi)/3` are co-terminal
Answers for `(0, 2pi):`
`pi/2; (2pi)/3, (4pi)/3, (3pi)/2`