Question

How do you find all solutions of the equation in the interval (0, 2pi) `2cos^2 theta+cos theta = 0`?

Answer 1

Factor out a cos.

Explanation:

`costheta(2costheta + 1) = 0`

`costheta = 0 or 2costheta + 1 = 0`

`theta = pi, (3pi)/2 or costheta = -1/2`

cos is negative in quadrants II and III. Applying the 30-60-90 special triangle and reference angles (leave a question on my homepage if you don't know how these work), we find that `theta = (2pi)/3 and (4pi)/3`

Thus, your solution set is `{theta = (3pi)/2, pi, (2pi)/3, (4pi)/3}`

Hopefully this helps!