Question #a6825

1 Answer
Feb 17, 2018

See below

Explanation:

Instead of #cos(2x)=0#, think of the equation as #cos(theta)=0#, where #theta=2x#

Since #costheta=0# at #theta=pi/2# and #theta=(3pi)/2#, then by substitution,

#cos(2x)=0# at #2x=pi/2# and #2x=(3pi)/2#

Solve each equation separately

#2x=pi/2#
#x=pi/2xx1/2#
#x=pi/4#

#2x=(3pi)/2#
#x=(3pi)/2xx1/2#
#x=(3pi)/4#