Question

How do you solve `2cos^2(theta) + sin(theta) = 1`?

Answer 1

`pi/2, (7pi)/6, and (11pi)/6`

Explanation:

`f(t) = 2cos^2 t + sin t - 1 = 0`
Replace `2cos^2 t` by `2(1 - sin^2 t)` -->
`f(t) = 2 - 2sin^2 t + sin t - 1 = 0`
`f(t) = - 2sin^2 t + sin t + 1 = 0`
Solve this quadratic equation for sin t.
Since a + b + c = 0, use shortcut. Two real roots:
sin t = 1 and `sin t = c/a = - 1/2`
Use trig table and unit circle -->
a. sin t = 1 -->`t = pi/2`
b. `sin t = - 1/2` --> There are 2 solution arcs t.
`t = - (5pi)/6` or `t = (7pi)/6` --> co-terminal arcs
and `t = pi - (-(5pi)/6)= (11pi)/6.`
Answers for `(0, 2pi):`
`pi/2, (7pi)/6 and (11pi)/6`