Question 788739
{{{csc(x)=1/sin(x)}}}does not exist when {{{sin(x)=0}}} at {{{x=0}}} and {{{x=pi}}}
For any other values of {{{x}}}, we can transform the equation into an equivalent quadratic equation by multiplying times {{{sin(x)}}}:
{{{3csc(x)-4sin(x)=0}}}<-->{{{3/sin(x)-4sin(x)=0}}}-->{{{3-4(sin(x))^2=0}}}
{{{3-4(sin(x))^2=0}}}-->{{{3=4(sin(x))^2=0}}}-->{{{(sin(x))^2=0= 3/4}}}
The solutions come from
{{{sin(x)=sqrt(3/4)=sqrt(3)/2}}} and
{{{sin(x)=-sqrt(3/4)=-sqrt(3)/2}}}
In the interval {{{"[0,"}}}{{{2pi}}}{{{")"}}}
{{{sin(x)=sqrt(3)/2}}}-->{{{system(highlight(x=pi/3), "or",highlight(x=2pi/3))}}} ( {{{sin(pi/3)=sin(60^o)=sqrt(3)/2}}} and being a supplementary angle {{{2pi/3=pi-pi/3}}} has the same sine)
and {{{sin(x)=-sqrt(3)/2}}}-->{{{system(highlight(x=4pi/6), "or",highlight(x=5pi/6))}}} (Those are co-terminal angles of {{{-pi/3}}} and {{{-2pi/3}}}, found adding {{{2pi}}} to {{{-pi/3}}} and {{{-2pi/3}}} )
 
NOTE:
For this kind of equation, a change of variable often helps to see the equation as a quadratic equation.
It is easier to see {{{3-4(sin(x))^2=0}}} as a quadratic equation using the change of variable {{{sin(x)=y}}}.
In this case, it was easy to solve the problem, even without mentioning a quadratic equation. In other cases, solving the quadratic may be more complivcated, so making the change of variable may save confusion and ink.
We get {{{3-4y^2=0}}}