Question 5027
{{{ 2 sin (x) + sqrt (3) = 0}}}
{{{ 2 sin (x) = - sqrt (3) }}}


Divide both sides by 2:
{{{sin (x) = - sqrt (3)/ 2}}}


{{{sin (pi/3) = sqrt (3) /2}}}, but in order for sin x to be negative, it must be in the third or fourth quadrant with a reference angle of {{{pi/3}}}.


Therefore, {{{ x = (4*pi)/3}}} {{{x= (5*pi)/3 }}}


R^2 at SCC