How do you solve #2sin2xcos2x#?

1 Answer
Nghi N.
Jun 21, 2018

#2sin 2x.cos 2x = sin 4x#.
To solve for x, it needs to have an equation.
For example, solve:
#2sin 2x.cos 2x = 0#
#sin 4x = 0#
Unit circle gives -->
#4x = 0 + 2kpi# --> #x = (kpi)/2#
#4x = pi + 2kpi = (2k + 1)pi# --># x = (2k + 1)pi/4#
#4x = (2pi) + 2kpi# = (k + 1)2pi --> #x = (k + 1)pi/2#

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