Consider the following simultaneous equations:

Question

3 cos^2 𝛼 + 2 cos^2 𝛽 =4
3 sin^2 𝛼 − 2sin^2 𝛽=0
*Find the values of cos 2𝛼 and cos 2𝛽.
*Hence solve for 𝛼𝛼 and 𝛽𝛽. Where 0°≤𝛼𝛼≤360° and 0°≤𝛽𝛽≤360°.

Answer 1

I'll just use cosx and cosy to avoid having to insert those Greek letters.

3cos^2x + 2cos^2y = 4
3sin^2x - 2sin^2y = 0
so,
3sin^2x = 2sin^2y
3-3cos^2x = 2-2cos^2y
2cos^2y = 3cos^2x - 1
that gives us
3cos^2x + 3cos^2x - 1 = 4
6cos^2x = 5
cos^2x = 5/6

cos^2x = 5/6
cos^2y = 3/4
Now just use your double angle formulas.