Question

How do you solve `9- 9\sin \theta = 6\cos^{2}theta`?

Answer 1

`pi/6, pi/2, (5pi)/6`

Explanation:

`9 - 9sin theta = 6cos^2 theta`
Replace `cos^2 theta` by `(1 - sin^2 theta)`
`9 - 9sin theta = 6(1 - sin^2 theta) = 6 - 6sin^2 theta`
or `6sin^2 theta - 9sin theta + 3 = 0`
Solve this quadratic equation for sin theta.
Since a + b + c = 0, use shortcut. There are 2 real roots:
`sin theta = 1` and `sin theta = c/a = 3/6 = 1/2`
Use trigonomteric table of special arcs and unit circle -->
a. `sin theta = 1` --> `theta = pi/2`
b. `sin theta = 1/2` --> `theta = pi/6`, and `theta = (5pi)/6`
Answers for (`0, 2pi)`
`pi/6, pi/2, (5pi)/6`