For #tan(x)=-1#
#color(white)("XXX")#Imagine a right triangle in standard position on the
#color(white)("XXX")#Cartesian plane.
#color(white)("XXX")tan(x)=-1# implies the vertical and horizontal sides
#color(white)("XXX")#sides of the right triangle are of equal magnitude
#color(white)("XXX")#but of opposite signs.
#color(white)("XXX")#This implies a reference angle of #pi/4# in Quadrant II or IV.
#color(white)("XXX")#In Quadrant II this would be #(3pi)/4#
#color(white)("XXX")#In Quadrant IV this would be #(3pi)/4+pi#
#color(white)("XXX")#Any addition of #pi# would simply generate a "new" angle
#color(white)("XXX")#superimposed on one of these two.
For #cos(2x)=1#
#color(white)("XXX")#Note that within the range #[0,2pi)# the only angle #theta#
#color(white)("XXX")# for which #cos(theta)=1# is #theta=0#
#color(white)("XXX")#So if #cos(2x)=1# in the range #[0,2pi)#
#color(white)("XXX")#then #2x=0 rArr x=0# OR #x=pi#
#color(white)("XXXXXXXXXX")#since #2pi# maps into the same angle as #0#
#color(white)("XXX")#Every addition of #pi# to these base angles will
map
#color(white)("XXX")#into a duplicate of one of these points.
