Question

How do you find the exact value of `2cos^2theta=sintheta+2` in the interval `0<=theta<360`?

Answer 1

`0; pi/6; (5pi)/6; pi; 2pi`

Explanation:

Replace `2cos^2 t = 2 - 2sin^2 t` in the equation:
`2 - 2sin^2 t = sin t + 2`
`- 2sin^2 t - sin t = 0`
sin t( -2sin t + 1) = 0
a. sin t = 0 --> arc `t = 0, t = pi, t = 2pi`
b. - 2sin t + 1 = 0 --> `sin t = 1/2`
Trig table of special arcs and unit circle give:
`sin t = 1/2` --> arc `t = pi/6` and `t = (5pi)/6`
Answers for (0, 2pi):
`0; pi/6; (5pi)/6; pi; 2pi`