trigonmetry help?

Question

heres the problem:

on the interval: 0 <(or equal) theta <(or equal) 2pi

2cos^2(theta) + cos(theta) = 0

i got this far:

cos(theta)*(2cos(theta)+1) = 0

cos(theta) = 0 (1st answer)

2nd answer:
2cos(theta) + 1 = 0
cos(theta) = -1/2 (2nd answer)

i know that cos 0 on the unit circle = pi/2
and cos 1/2 = pi/3

Answer 1

0 <= t < 2pi

2cos^2(t) + cos(t) = 0

cos(t) [ 2cos(t) + 1 ] = 0

Which branches off into these solutions:

cos(t) = 0
cos(t) = -1/2

cos(t) is equal to 0 at the points pi/2 and 3pi/2, so
t = {pi/2, 3pi/2}

cos(t) is equal to -1/2 at the points 2pi/3 and 4pi/3, so
t = {2pi/3, 4pi/3}

Therefore,

t = {pi/2, 3pi/2, 2pi/3, 4pi/3}

Remember that, due to the interval restriction from 0 to 2pi, you're going to (for the most part) always get two solutions on your unit circle, with the exceptions being if sin(x) = +/- 1 or cos(x) = +/- 1, which will give you one solution.

Answer 2

You are missing angles because cos t = 0 gives t=pi/2 and t= 3pi/2
and cos t =-1/2 gives t=2pi/3 and t= 4pi/3

Answer 3

You are correct!
but since cos(theta) is negative, the angle has to be between pi and 2 pi - (4pi/3 for example, or (5pi/3)

Answer 4

cos = 0 at pi/2, 3pi/2
cos = -1/2 at 2pi/3, 4pi/3

and the full unit circle can be found on
http://en.wikipedia.org/wiki/Unit_circle