Question

How do you solve `Cos (2theta) + 6sin^2 theta = 4`?

Answer 1

`pi/3, (2pi)/3, (4pi)/3, (5pi)/3`

Explanation:

Use trig identity:
`cos 2t = 1 - 2sin^2 t.`
The equation is transformed to:
`1 - 2sin ^2t + 6sin^2 t = 4`
`4sin^2 t = 3`
`sin^2 t = 3/4`
`sin t = +- sqrt3/2`

Trig table and trig unit circle -->
a. `sin t = sqrt3/2` --> `t = pi/3` and `t = (2pi)/3`
b. `sin t = - sqrt3/2` --> `t = (4pi)/3` and `t = (5pi)/3`