# 2sinxcosx-sin(2x)cos(2x)=0. Simplify the left side of the expression so that each term involves only 2x, factor the left side so that it can be solved.Also, Once i get the factored equation how...

2sinxcosx-sin(2x)cos(2x)=0. Simplify the left side of the expression so that each term involves only 2x, factor the left side so that it can be solved.

Also, Once i get the factored equation how would I solve it?

Thank you so much! Any help is greatly appreciated!

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Solve `2sinxcosx-sin(2x)cos(2x)=0`

Use the trig identity `sin(2x)=2sinxcosx` :

2sinxcosx-sin(2x)cos(2x)=0

sin(2x)-sin(2x)cos(2x)=0

sin(2x)(1-cos(2x))=0

So by the zero product property either sin(2x)=0 or 1-cos(2x)=0

(a) sin(2x)=0 The sine is 0 at `x=+-kpi` where k is an integer. So sin(2x)=0 ==> `x=pi/2+-kpi/2`

** Or `2x=sin^(-1)(0)==>2x=0+-kpi ==> x=0+-kpi/2` which is equivalent **

(b) 1-cos(2x)=0 ==>cos(2x)=1 Then

`2x=cos^(-1)(1) ==>2x=0+-2pi ==>x=0+-pi`

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The solutions to 2sinxcosx-sin(2x)cos(2x)=0 are `x=0+-kpi/2`

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** Note that integral multiples of `pi` are multiples of `pi/2` **

The graph of y=2sinxcosx-sin(2x)cos(2x):

sorry, which belongs to which? As in what problem fits with what I asked I got confused. thanks