Question

How would I go about solving this equation on the interval [0,2pi]

cos^2 + 2 cos x + 1 = 0

Answer 1

cos^2(x) + 2 cos(x) + 1 = [cos(x) + 1]^2

So, cos(x) = -1 ------>

x = (2n+1)pi

Answer 2

Muliple choice questions I have 4 options

answers.
1. pi/4, 7pi/4
2. pi/2, 3pi/2
3. pi
4. 2pi

What formula do I use ?

Answer 3

thanks