Question

How do you solve `sin (x/2) + cosx=1`?

Answer 1

`0, 2pi, pi/3, and (5pi)/3`

Explanation:

Apply the trig identity: `cos x = 1 - 2sin^2 (x/2)`
`sin (x/2) + 1 - 2sin^2 (x/2) - 1 = 0`
`sin (x/2) - 2sin^2 (x/2) = sin (x/2)(1 - 2sin (x/2)) = 0`
a. `sin (x/2) = 0` --> 3 solutions -->
`x/2 = 0` --> `x = 0`
`x/2 = pi` -->`x = 2pi`
`x/2 = 2pi` --> `x = 4pi`
b. `sin (x/2) = 1/2` --> 2 solutions -->
`x/2 = pi/6` --> `x = (2pi)/6 = pi/3`
`x/2 = pi - pi/6 = (5pi)/6` --> `x = (5pi)/3`