Question 436994
As you know {{{(sin(x))^2=1-(cos(x))^2}}}, substitute in our equation and have:

{{{2(1-(cos(x))^2)+3(cos(x))-3=0}}} => {{{-2(cos(x))^2+3(cos(x))-1=0}}} =>

{{{2(cos(x))^2-3cos(x)+1=0}}}, substitute cos(x)=y and write:

{{{2y^2-3y+1=0}}},solving this quadratic equation have: y=1 and y=1/2.

We now solve two new equivalent equations: 1)cos(x)=1 and 2) cos(x)=1/2,

1) {{{cos(x)=1}}} => {{{x=2n*pi}}}.

2) {{{cos(x)=1/2}}} => {{{x=2n*pi+(pi/3)}}} and {{{x=2n*pi+(5*pi/3)}}},where n 

positive integer.