2(Sinx)^2 + Cosx - 2 = 0.
Recall (Sinx)^2 + (Cosx)^2 = 1.
Therefore (Sinx)^2 = 1 - (Cosx)^2
Substitute this into the question above.
2(Sinx)^2 + Cosx - 2 = 0.
2(1 - (Cosx)^2) + Cosx - 2 = 0 Expand
2 - 2(Cosx)^2 + Cosx - 2 = 0
2 - 2 - 2(Cosx)^2 + Cosx = 0
- 2(Cosx)^2 + Cosx = 0 Multiply both sides by -1.
2(Cosx)^2 - Cosx = 0
Let p = Cosx
2p^2 - p = 0 Factorise
p(2p - 1) = 0. Therefore p=0 or (p-1) = 0
p=0 or (p-1) = 0
p=0 or p = 0 +1.
p=0 or p = 1 Recall p = Cosx
Therefore Cosx = 0 or 1.
For 0<x<360
Cosx = 0, x = Cos inverse (0) , x = 90, 270
Cosx = 1, x = Cos inverse (1) , x = 0, 360
Therefore x = 0,90, 270 & 360 degrees.
Cheers.
