SOLUTION: solve cos^2x-3sinx=3 where 0 < x < / 2 pi

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Question 878370: solve cos^2x-3sinx=3 where 0 < x < / 2 pi
Answer by lwsshak3(11628) About Me  (Show Source):
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solve cos^2x-3sinx=3 where 0 < x < / 2 pi
cos^2(x)-3sinx=3
1- sin^2(x)-3sinx-3=0
- sin^2(x)-3sinx-2=0
sin^2(x)+3sinx+2=0
(sinx+2)(sinx+1)=0
..
sinx+2=0
sinx=-2 (reject(-1 ≤ x ≤1))
or
sinx+1=0
sinx=-1
x=3π/2