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
- 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
Exact solutions in the interval 0 <= θ <= 2*pi:
θ = arccos(-1/3) or θ = 2*pi - arccos(-1/3)