Question 830596
[0,2pi]: sin^2(x)-sq root of 3 sin(x)+1=cos^2(x)
..
{{{sin^2(x)-sqrt(3)sin(x)+1=cos^2(x)}}}
{{{sin^2(x)-sqrt(3)sin(x)+1=1-sin^2(x)}}}
{{{2sin^2(x)-sqrt(3)sin(x)=0}}}
{{{sin(x)(2sin(x)-sqrt(3))=0}}}
..
sin(x)=0
x=0,π
or
2sin(x)-√3=0
sin(x)=√3/2
x=π/3,2π/3 (in quadrants I and II in which sin>0.