We must use following properties:
#color(red)(sin x = sin(2npi+x))#
#color(red)(sin x = sin((2n-1)pi-x))#
Where #n# is an integer.
Since #sqrt3/2 >0#, #theta# has to be in the first and second quadrant #(0<=theta<=pi# and #2pi<=theta<=3pi)#.
We know #sin (color(red)(pi/3))= sqrt3/2# so using the first identity we get:
#sin (pi/3) = sin (2pi+pi/3)#
#sin (pi/3) = sin (color(red)((7pi)/3))#
And using the second:
#sin (pi/3)= sin(pi-pi/3)#
#sin (pi/3)=sin (color(red)((2pi)/3))#
First one again:
#sin ((2pi)/3)=sin(2pi+(2pi)/3)#
#sin ((2pi)/3)=sin (color(red)((8pi)/3))#
We can only use #n=1# because if it would to be bigger, #theta > 4pi#, which is not what we need.
All the values in red are the solutions to the equality
#sin theta = sqrt3/2# ,
so #theta -> {color(red)(pi/3),color(red)((2pi)/3),color(red)((7pi)/3),color(red)((8pi)/3)}#.
