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
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.