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Evgesh-ka [11]
3 years ago
10

Solve 3-2cos²x-3sinx=0 for 0≤x≤360

Mathematics
1 answer:
crimeas [40]3 years ago
7 0
<span>3-2(Cosx)^2 - 3Sinx = 0.

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

</span><span>3-2(Cosx)^2 - 3Sinx = 0
3 - 2(1 - (Sinx)^2)  - 3Sinx = 0  Expand
3 - 2 + 2(Sinx)^2  -  </span><span><span>3Sinx = 0</span>
1 + 2(</span><span>Sinx)^2  - 3Sinx  = 0  Rearrange
2(Sinx)^2 </span><span><span>- 3Sinx + </span>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<u><</u>x<u><</u>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.</span>
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