Question

How do you solve `cos2theta + 6sin^2theta = 2`?

Answer 1

θ is 30 deg or 120 deg

Explanation:

#cos2θ + 6sin²θ = 2 or, cos²θ − sin²θ + 6sin²θ = 2 or, cos²θ + sin²θ + 4sin²θ = 2 or, 1 + 4sin²θ = 2 or, sin²θ = 1/4 or, sinθ = +-1/2 or, θ = 30 deg, 120 deg#

Answer 2

`pi/6, (5pi)/6;(7pi)/6; (11pi)/6`

Explanation:

`cos 2t + 6sin^2 t = 2`
Replace `cos 2t` by `(1 - 2sin^2 t)` --> trig identity:
`1 - 2sin^2 t + 6sin^2 t = 2`
`4sin^2 t = 1`
`sin^2 t = 1/4`
`sin t = +- 1/2`
Trig table and unit circle give 4 solutions:
a. `sin t = 1/2` -->
`t = pi/6 + 2kpi`, and
`t = (5pi)/6 + 2kpi`
b. `sin t = - 1/2` -->
`t = - pi/6`, or, as co-terminal, `t = (11pi)/6 + 2kpi`
`t = pi - (-pi/6) = (7pi)/6 + 2kpi`