Question

How do you solve `cos^2x+2cos2x+1=0`?

Answer 1

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

Ans: `+- 63.43` and `+- 116.57` deg

Explanation:

Replace in the equation `cos 2x` by `(2cos^2 x - 1)`
`f(x) = cos^2 x + 4cos^2 x - 2 + 1 = 0`
`5cos^2 x - 1 =`
`cos^2 x = 1/5` --> `cos x = +- 1/sqrt5 = +- 0.48`

a. `cos x = 0.48` --> `x = +- 63.43` deg

b. `x = - 0.48` --> `x = +- 116.57`
Check by calculator:
x = 63.43 deg -->`cos^2 x = 0.20` ; 2cos 2x = - 1.20.
f(x) = 0.20 - 1.20 + 1 = 0. OK
x = 116.57 --> `cos^2 x = 0.20` ; 2cos 2x = - 1.20.
f(x) = 0.20 - 1.20 + 1 = 0. OK