SOLUTION: solve for x-values between [0,2pi) 6sin^2x+9sinx+3=0

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Question 822038: solve for x-values between [0,2pi)
6sin^2x+9sinx+3=0
Found 2 solutions by KMST, lwsshak3:
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
6sin%5E2%28x%29%2B9sin%28x%29%2B3=0
2sin%5E2%28x%29%2B3sin%28x%29%2B1=0
%282sin%28x%29%2B1%29%29%28sin%28x%29%2B1%29=0
Either
2sin%28x%29%2B1=0-->2sin%28x%29=-1-->sin%28x%29=-1%2F2 ,
or
sin%28x%29%2B1=0-->sin%28x%29=-1 .

system%28sin%28x%29=-1%2F2%2C0%3C=x%3C2pi%29 --> system%28highlight%28x=11pi%2F6%29%2C+%22or%22%2Chighlight%28x=7pi%2F6%29%29
because sin%28pi%2F6%29=1%2F2-->sin%28-pi%2F6%29=-1%2F2-->sin%282pi-pi%2F6%29=-1%2F2-->sin%2811pi%2F6%29=-1%2F2 and
sin%28pi%2F6%29=1%2F2-->sin%28pi%2Bpi%2F6%29=-1%2F2-->sin%287pi%2F6%29=-1%2F2

system%28sin%28x%29=-1%2C0%3C=x%3C2pi%29 --> highlight%28x=pi%29

TIP:
When you get a problem like, you can change variable, or at least think of it that way.
In this case, if you substitute sin%28x%29=y the equation transforms into the simple quadratic equation
6y%5E2%2B9y%2B3=0 .
It is a lot easier to write, and a lot easier to see how to factor it and solve it.

Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
solve for x-values between [0,2pi)
6sin^2x+9sinx+3=0
divide by 3
2sin^2x+3sinx+1=0
factor:
(2sinx+1)(sinx+1)=0
..
2sinx+1=0
sinx=-1/2
x=5π/4,7π/4 (in quadrant III and IV in which sin>0)
or
sinx+1=0
sinx=-1
x=3π/2