Question 822038
{{{6sin^2(x)+9sin(x)+3=0}}}
{{{2sin^2(x)+3sin(x)+1=0}}}
{{{(2sin(x)+1))(sin(x)+1)=0}}}
Either
{{{2sin(x)+1=0}}}-->{{{2sin(x)=-1}}}-->{{{sin(x)=-1/2}}} ,
or
{{{sin(x)+1=0}}}-->{{{sin(x)=-1}}} .
 
{{{system(sin(x)=-1/2,0<=x<2pi)}}} --> {{{system(highlight(x=11pi/6), "or",highlight(x=7pi/6))}}}
because {{{sin(pi/6)=1/2}}}-->{{{sin(-pi/6)=-1/2}}}-->{{{sin(2pi-pi/6)=-1/2}}}-->{{{sin(11pi/6)=-1/2}}} and
{{{sin(pi/6)=1/2}}}-->{{{sin(pi+pi/6)=-1/2}}}-->{{{sin(7pi/6)=-1/2}}}
 
{{{system(sin(x)=-1,0<=x<2pi)}}} --> {{{highlight(x=pi)}}}
 
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(x)=y}}} the equation transforms into the simple quadratic equation
{{{6y^2+9y+3=0}}} .
It is a lot easier to write, and a lot easier to see how to factor it and solve it.