To find the range, proceed as follows :
#-1 <= sinx <=1#
#-1 <= sin2x <=1#
#1 >= -sin2x >=-1#
#2 >= -2sin2x >=-2#
#2+5 >= (5-2sin2x )>=-2+5#
#3<= f(x)=(5-2sin2x) <=7#
Therefore,
the range is #f(x) in [3,7]#
To sketch the graph in the domain #x in (0, pi)#
Calculate the following values
#color(white)(aaaa)##x##color(white)(aaaa)##f(x)#
#color(white)(aaaa)##0##color(white)(aaaa)##5#
#color(white)(aaaa)##pi/4##color(white)(aaaa)##3#
#color(white)(aaaa)##pi/2##color(white)(aaaa)##5#
#color(white)(aaaa)##3/4pi##color(white)(aaaa)##7#
#color(white)(aaaa)##pi##color(white)(aaaa)##5#
#5-2sin2x=6#
#2sin2x=-1#
#sin2x=-1/2#
#2x=7pi/6#, #=>#, #x=7/12pi#
#2x=11/6pi#, #=>#, #x=11/12pi#
See the graph below
graph{(y-5+2sin(2x))(y-6)=0 [-10, 10, -5, 5]}