Question 1011622

6cos^2(x)+3sin(x)=3
for (0 < x < 360)

I've got to solution:
sin(x)-1=0
x=90

Or
2sin(x)+3=0
sin(x)=-3/2
The second solution confuses me because I can't take arc sin -3/2. I must have made a mistake somewhere. Any help please?
<pre>Solve {{{2 cos^2 (x) - sin x - 1 = 0}}} ---------> (2 sin x + 1)(sin x - 1) = 0, to get:
{{{highlight_green(system(matrix(1,3,sin (x) = - 1/2, or, 210^o)))}}} (3<sup>rd</sup> quadrant)
{{{highlight_green(system(matrix(1,3,sin (x) = - 1/2, or, 330^o)))}}} (4<sup>th</sup> quadrant)
{{{highlight_green(system(matrix(1,3,sin (x) = 1, or, 90^o)))}}} (Positive side of y-axis)