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solve 3-2cos²x-3sinx=0 for 0≤x≤360

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Olemakpaduidn
3-2(Cosx)^2 - 3Sinx = 0.

Recall (Sinx)^2 + (Cosx)^2  = 1.
Therefore  (Cosx)^2 =  1 - (Sinx)^2
Substitute this into the question above.

3-2(Cosx)^2 - 3Sinx = 0
3 - 2(1 - (Sinx)^2)  - 3Sinx = 0  Expand
3 - 2 + 2(Sinx)^2  -  3Sinx = 0
1 + 2(Sinx)^2  - 3Sinx  = 0  Rearrange
2(Sinx)^2 - 3Sinx + 1  = 0 
Let p = Sinx
2p^2  -  3p + 1  = 0    Factorise the quadratic expression
2p^2 - p - 2p +1 = 0
p(2p -1) - 1(2p -1) = 0
(2p-1)(p -1) = 0

Therefore  2p-1=0    or  (p-1) = 0
2p=0+1    or  (p-1) = 0
2p=1       or    p = 0 +1.
p=1/2       or  p = 1            Recall p = Sinx

Therefore Sinx = 1/2 or 1.
For  0<x<360

Sinx =1/2,  x = Sin inverse (1/2) , x = 30, 
                                   (180-30)- 2nd Quadrant = 150 deg
Sinx = 1,  x = Sin inverse (1) , x = 90

Therefore x = 30,90 & 150 degrees.

Cheers.
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