# I have a single trig problem I do not understand, could you please explain how I can solve this? ?

Here's the question

Solve algebraically cos(θ)+2cos(θ)+1=0 with exact solutions in the interval [0, 2π)

A walkthrough would be really apricated

### 2 Answers

- az_lenderLv 71 month agoFavorite Answer
Really? Are you sure you didn't mean cos^2(theta) + 2*cos(theta) + 1 = 0 ?

The equation you actually wrote means 3cos(theta) = -1, so

theta = arccos(-1/3), one solution in the 2nd quadrant and one in the 3rd quadrant, but neither of them any rational number of degrees.

If you meant cos^2(theta) + 2*cos(theta) + 1 = 0, then it's the same as

[cos(theta) + 1] ^ 2 = 0, so

cos(theta) = -1, which implies theta = pi.

- Jeff AaronLv 71 month ago
cos(θ) + 2*cos(θ) + 1 = 0

3*cos(θ) + 1 = 0

3*cos(θ) = -1

cos(θ) = -1 /3

General solution (exact):

θ = 2*pi*n +/- arccos(-1/3), for any integer n

Approximate:

https://www.wolframalpha.com/input/?i=cos%28%CE%B8...

Exact solutions in the interval 0 <= θ <= 2*pi:

θ = arccos(-1/3) or θ = 2*pi - arccos(-1/3)