3 cos^2 ๐ผ + 2 cos^2 ๐ฝ =4
3 sin 2 ๐ผ โ 2sin 2 ๐ฝ=0
Find the values of cos 2๐ผ and cos 2๐ฝ.
I already did this for you. Take the trouble to read it this time, ok?
since sin^2 + cos^2 = 1, the 2nd equation becomes
3(1-cos^2 ๐ผ) โ 2(1-cos^2 ๐ฝ) = 0
3 - 3cos^2 ๐ผ โ 2 + 2cos^2 ๐ฝ = 0
Now the two equations read
3cos^2 ๐ผ + 2cos^2 ๐ฝ = 4
-3cos^2 ๐ผ + 2cos^2 ๐ฝ = -1
subtract to get
6cos^2๐ผ = 5
cos^2๐ผ = 5/6
Now just finish solving for cos^2๐ผ and cos^2๐ฝ
Then use your double-angle formula
cos 2x = 2cos^2x - 1
post your work if you get stuck, but don't just post the same problem all over again.