Question 1019974
The left side of that equation should remind you of the trigonometric identity
{{{sin(A+B)=sin(A)cos(B)+cos(A)sin(B)}}} .
Making {{{system(A=2x,B=x)}}} (or the other way around),
{{{sin(2x+x)=sin(2x)cos(x)+cos(2x)sin(x)}}} , or
{{{sin(2x)cos(x)+cos(2x)sin(x)=sin(3x)}}} .
So, we can re-write the equation as
{{{sin(3x)=-1/2}}} .
In the {{{"["}}}{{{-pi}}}{{{" ,"}}}{{{pi}}}{{{"]"}}} interval, the solutions for {{{3x}}} are
{{{3x=-pi/6}}} or {{{3x=-5pi/6}}} .
We could write that as {{{3x=-pi/2 +- pi/3}}} .
Since sine has a period of {{{2pi}}} ,
other values for {{{3x}}} differing in any multiple of {{{2pi}}} also have {{{sin(3x)=-1/2}}} .
All the solutions can be written as
{{{3x=-pi/2 +-pi/3 +2k*pi=(4k-1)pi/2 +- pi/3}}} for any integer {{{k}}} .
So, {{{highlight(x=(4k-1)*pi/2 +- pi/9)}}} for any integer {{{k}}} , or
{{{highlight(x=(12k-3 +- 2)*pi/18)}}} for any integer {{{k}}} .