Question

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

Answer 1

`0, 2pi, pi/3; 5pi/3`-->interval `(0, 2pi)`

Explanation:

Use trig identity: `cos 2a = 1 - 2sin^2 a`
`cos x = 1 - 2sin^2 (x/2)`.
Call `sin (x/2) = t`, we get:
`t + (1 - 2t^2) - 1 = t(1 - 2t) = 0`
2 solutions:
a. `t = sin (x/2) = 0`-->
`x/2 = 0` --> `x = 0`, and
`x/2 = pi` --> `x = 2pi`
b. 1 - 2t = 0 --> `t = sin x = 1/2`
`t = sin x/2 = 1/2` --> 2 answers:
`x/2 = pi/6` --> `x = pi/3`
`x/2 = (5pi)/6 --> x = (5pi)/3`
Check
x = pi/3 --> x/2 = pi/6 --> sin x/2 = 1/2 --> cos pi/3 = 1/2 -->
1/2 + 1/2 - 1 = 0. Correct
x = (5pi)/3 --> x/2 = (5pi)/6 --> sin (5pi)/6 = 1/2 --> cos (5pi)/3 = cos (pi/3) = 1/2 --> 1/2 + 1/2 - 1 = 0. OK