#4cos(x)+5 = 3#
#=> 4cos(x) = -2#
#=> cos(x) = -1/2#
From a unit circle, we can tell that #cos(x) = -1/2# at #x = +-pi/3+2pin, n in ZZ#. Note that we add the #2pin# as #cos(x)# is periodic with a period of #2pi#.
#=> x = pi/3 + 2pin or x = -pi/3 + 2pin#
Now we can look to see what values for #n# leave #x# within our desired interval of #[0, 2pi]#
For #x = pi/3+2pin#, only #n=0# works.
For #x = -pi/3+2pin#, only #n=1# works.
Thus, we get our final solution set
#x = pi/3+2pi*0 = pi/3#
or
#x = -pi/3 + 2pi*1 = (5pi)/3#
#:. x in {pi/3, (5pi)/3}#
