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

Question 822038: solve for x-values between [0,2pi)
6sin^2x+9sinx+3=0
Found 2 solutions by KMST, lwsshak3:
Answer by KMST(5328) (Show Source): You can put this solution on YOUR website!

Either
-->--> ,
or
--> .

-->
because -->-->--> and
-->-->

-->

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 the equation transforms into the simple quadratic equation
.
It is a lot easier to write, and a lot easier to see how to factor it and solve it.

Answer by lwsshak3(11628) (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

RELATED QUESTIONS
Solve 8cos^2(x)-6sin(x)-9 = 0 for all solutions 0 <= x <... (answered by lwsshak3)
Solve the trigonometric equation, using exact values when possible, where x is a real... (answered by jim_thompson5910)
Solve sin 2x = sin x for x between 0 and 2pi. (answered by stanbon)
how do i solve the following equation for values of x in the range 0 degrees < x <360... (answered by Edwin McCravy)
6sin^2x - sinx - 2 =... (answered by AnlytcPhil)
6sin^2x-13sinx-8=0 (answered by Alan3354)
Find all solutions on the interval x element [0, 2pi) for the trigonometric equation... (answered by Fombitz)
sin (2x) = cos (2x) between (0,... (answered by Alan3354)
I need to solve this equation for 0 < x< 2pi -2tan(-2x + 4pi/3) = 2sqrt3 (answered by stanbon,ikleyn)