Question 448400
{{{2(sinx)^2-sqrt(3)*sinx=0}}}, assuming sinx≠0 or x≠n*{{{pi}}},

divide both sides by sinx and get:{{{2sinx=sqrt(3)}}}=>{{{sinx=sqrt(3)/2}}}, 

where its solution is:{{{x=2n*pi+(pi)/3}}} and {{{x=2n*pi+2*pi/3}}}.