Question

How do you find the exact value of `cos2theta+2cos^2theta=2` in the interval `0<=theta<360`?

Answer 1

The solutions are `theta={30º,150º,210º,330º}`

Explanation:

We use `cos2theta=2cos^2theta-1`

Therefore our equation is

`cos2theta+2cos^2theta-2=0`

`2cos^2theta-1+2cos^2theta-2=0`

`4cos^2theta=3`

`cos^2theta=3/4`

`cos theta=+-sqrt3/2`

First, `costheta=sqrt3/2`, `=>`, `theta=30º ; 330º`

Second, `costheta=-sqrt3/2`, `=>`, `theta=150º; 210º`

Answer 2

Please see the explanation

Explanation:

Given: `cos(2theta) + 2cos^2(theta) = 2; 0^@ le theta < 360^@`

The identity `cos(2theta) = 2cos^2(theta) - 1` allows allows us to substitute `2cos^2(theta) - 1` for `cos(2theta)`:

`2cos^2(theta) - 1 + 2cos^2(theta) = 2`

Combine like terms:

`4cos^2(theta) = 3`

Divide by 4:

`cos^2(theta) = 3/4`

square root both sides:

`cos(theta) = +-sqrt(3)/2`

The angles are well known:

`theta = 30^@, 150^@, 210^@, and 330^@`