Question 752819
{{{sin(x)+sqrt(3)cos(x)=0}}} --> {{{sin(x)=-sqrt(3)cos(x)}}} --> {{{sin(x)/cos(x)=-sqrt(3)}}} --> {{{tan(x)=-sqrt(3)}}}
We know that {{{tan(60^o)=sqrt(3)}}}
So {{{tan(-60^o)=-sqrt(3)}}} because {{{tan(-A)=-tan(A)}}}
and also {{{tan(180^o-60^o)=tan(120^o)=-sqrt(3)}}} because {{{tan(A+180^o)=tan(A)}}}
So {{{highlight(x=-60^o)}}} and {{{highlight(x=120^o)}}} are two solutions,
and in general
{{{highlight(x=k*180^o-60^o)}}} for all integer values of {{{k}}} represents all the solutions.
 
If you need to measure angles in radians, it would be
{{{highlight(x=k*pi-pi/3)}}}