2sinxcosx - sin(2x)cos(2x) = 0
Part I
The double angle identity for sine states that sin(2x) = 2sinxcosx
Thus we get:
sin(2x) - sin(2x)cos(2x) = 0
Part II
sin(2x)(1 - cos(2x)) = 0
Part III
Either sin(2x) = 0 or
1 - cos(2x) = 0
=> cos(2x) = 1
For sin(2x) = 0, this is true for
2x = n(pi) where n = 0, 1, 2, ....
x = n(pi/2)
For cos(2x) = 1, this is true for
2x = n(pi) where n = 0, 2, 4, ....
x = n(pi/2)
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