Question

How do you find the exact value of `2sin^2theta+3costheta-3=0` in the interval `0<=theta<360`?

Answer 1

`0, 60^@, 300^@, 360^@`

Explanation:

Replace in the equation `sin^2 t` by `(1 - cos^2 t)` -->
`2 - 2cos^2 t + 3cos t - 3 = 0`
`- 2cos^2 t + 3cos t - 1 = 0`
Solve this quadratic equation for cos t.
Since a + b + c = 0, use shortcut. The 2 real roots are:
cos t = 1 and `cos t = c/a = (-1)/(-2) = 1/2`

Use trig table and unit circle -->
a. cos t = 1 --> arc t = 0 and t = `2pi`
b. `cos t = 1/2` --> arc `t = +- pi/3`
The co-terminal arc of (-pi/3) is (5pi)/3
Answers for `(0, 2pi)`:
`0, pi/3, (5pi)/3, 2pi`, or
`0, 60^@, 300^@, 360^@`