Question

How do you solve `cos 2(x)- cos(x)- 1 = 0`?

Answer 1

`x = +- 141^@26 + k360^@`

Explanation:

f(x) = cos 2x - cos x - 1 = 0
Replace cos 2x by `(2cos^2 x - 1)` -->
`f(x) = 2cos^2 x - cos x - 2 = 0`
Solve this quadratic equation for cos x -->
`D = b^2 - 4ac = 1 + 16 = 17` --> `d = +- sqrt17`
There are 2 real roots:
`cos x = -b/(2a) +- d/(2a) = 1/4 +- sqrt17/4 = (1 +- sqrt17)/4`
a. `cos x = (1 + 4.12)/4 = 5.12/4 = 1.28` (rejected because > 1)
b. `cos x = (1 - 4.12)/4 = - 3.12/4 = - 0.78`
Calculator and unit circle give -->
`x = +- 141^@26 = k360^@`
Check by calculator.
`x = 141^@26` --> cos x = - 0.78 --> `2x = 282^@52` --> cos 2x = 0.22 -->
f(x) = 0.22 - (- 0.78) - 1 = 0. Proved.