Question

How do you solve `3- 3 Cos (theta) = 2 Sin^2 (theta)`?

Answer 1

In `[0, 2pi], theta = 0, pi/3, 5pi/3 and 2pi`.
General solution: `theta=2kpi and theta= 2kpi+-pi/3`, k=1,2,3,....

Explanation:

Using `sin^2theta=1-cos^2theta and c=costheta`,,
`3-3c=2(1-c^2)`
`2c^2-3c+1=0`
`c=cos theta=1, 1/2`..

In `[0, 2pi]: arc cos 1=0 and 2pi`.
#arc cos(1/2)=pi/3 and (5pi)/3#,
using `cos(2pi+-theta)=costheta`

General solution is given by `cos(2kpi+-theta)=costheta`, k=1,2,3....