Alternate answer, using the quadratic formula, as requested.
First, find where the left hand side is equal to #0# using the quadratic formula:
#(sinx-1/2)(sinx-2) = 0#
#=> (sinx)^2-5/2sinx + 1 = 0#
#=> sinx = (-(-5/2)+-sqrt((-5/2)^2-4(1)(1)))/(2(1))#
#=(5/2+-sqrt(9/4))/2#
#=(5/2+-3/2)/2#
#=5/4+-3/4#
#=> sinx = 2 or sinx = 1/2#
(Note that we could have arrived at this conclusion directly from the factored form by setting each factor equal to #0#)
As #sinx != 2# for any #x in RR#, this leaves us with
#sinx = 1/2#
Using knowledge of common angles, a unit circle, or a calculator, we find that #sinx = 1/2# has the solutions #x in {pi/6, (5pi)/6}# in the interval #[0, 2pi]#.
As #(sinx-1/2)(sinx-2)# is a continuous function, any interval in which it changes sign must include a #0#. Thus, if we split the interval #[0, 2pi]# by removing the points #{pi/6, (5pi)/6}#, we get three intervals:
#[0, pi/6)#, #(pi/6, (5pi)/6)#, #((5pi)/6, 2pi]#
and the sign of #(sinx-1/2)(sinx-2)# is constant on each of these intervals, as none of them contain a #0# of #(sinx-1/2)(sinx-2)#.
With that, we can determine the sign of #(sinx-1/2)(sinx-2)# on each interval by testing a value in each one, and noting that the sign of that value will be the same as the sign of all of the values on that interval.
#0 in [0, pi/6)# and
#(sin(0)-1/2)(sin(0)-2) = (-1/2)(-2) = 1 > 0#
Thus #(sinx-1/2)(sinx-2) > 0# on #[0, pi/6)#
#1 in (pi/6, (5pi)/6)# and
#(sin(1)-1/2)(sin(1)-2) = (1/2)(-1) = -1/2 < 0#
Thus #(sinx-1/2)(sinx-2) < 0# on #(pi/6, (5pi)/6)#
#pi in ((5pi)/6, 2pi]# and
#(sin(2pi)-1/2)(sin(2pi)-2) = (-1/2)(-2) = 1 > 0#
Thus #(sinx-1/2)(sinx-2) > 0# on #((5pi)/6, 2pi]#
Adding in the points at which the function is #0#, we obtain our final result:
#(sinx-1/2)(sinx-2) <= 0# if #x in [pi/6, (5pi)/6]#
