Find Φ given that 2cos^2Φ=3+5cosΦ and 0<Φ<360
2cos^2Ø - 5cosØ - 3 = 0
(2cosØ+1)(cosØ-3) = 0
cosØ = -1/2 or 3
since |cosØ| <= 1, the only solution is
cosØ = -1/2, so
Ø = 120° or 240°
(2cosØ+1)(cosØ-3) = 0
cosØ = -1/2 or 3
since |cosØ| <= 1, the only solution is
cosØ = -1/2, so
Ø = 120° or 240°