Question

How do you find the general solutions for `cos(x/2) + cosx - 1 = 0`?

Answer 1

Solve `f(x) = cos (x/2) + cos x - 1 = 0`

Ans: 77.48 + k360 deg

Explanation:

Call `cos (x/2) = t`
Use trig identity: `cos x = 2t^2 - 1.`
`f(x) = t + 2t^2 - 2 = 2t^2 + t - 2`.
`D = d^2 = b^2 - 4ac = 1 + 16 = 17` --> `d = +- sqrt17 = +- 4.12`

`t1 = -1/4 + sqrt17/4 = 3.12/4 = 0.78` (accepted)
`t2 = -1/4 - sqrt17/4 = -5.12/4 = - 1.28`(rejected)
`cos (x/2) = 0.78` --> `x/2 = 38/74` deg --> `x = 77.48` deg
General answer: x = 77.48 + k360.
Check.
x = 77.48 --> cos x = 0.22 --> x/2 = 38.74 --> cos (x/2) = 0.78.
cos (x/2) + cos x - 1 = 0.78 + 0.22 - 1 = 0. OK