Solve 2sin²x+cosx-2=0 for o≤x≤360

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.

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