Question

How do you solve `cos^2x+2cosx+1=0` over the interval 0 to 2pi?

Answer 1

Solve as a quadratic first to find the value of `cos(x)`.

Explanation:

Factorize the left hand side.

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

This means that

`1 + cos(x) = 0`

or

`cos(x) = -1`

From the graph of `y = cos(x)`
graph{cos(x) [-10, 10, -5, 5]}
The only value of `x` in the interval `0 <= x <= 2pi` that gives `cos(x) = -1` is `x = pi`.