Solve the equation #(sin^2x)(sin2x)=0# where #0lexle2pi#?

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Answer 1 · 2018-04-14T05:04:12

#x=0, pi, pi/2, (3pi)/2#

We may solve the separate equations #sin2x=0, sin^2x=0#:

For #sin2x=0,# in the interval #[0, 2pi], # we have the solutions

#2x=0, 2x=2pi#

Solve for #x:#

#x=0/2=0#

#x=2pi/2=pi#

Thus, #x=0, pi# are our solutions so far.

For #sin^2x=0,# recall the identity #sin^2x=1-cos^2x#. We then solve:

#1-cos^2x=0#

#cos^2x=1#

#cosx=+-sqrt1#

#cosx=+-1#

In the interval #[0, 2pi],# the solutions are #x=pi/2, (3pi)/2#