# Precalculus

sin^2(2x)=2sinxcosx. Find all solutions to each equation in the interval [0, 2pi)

So I started off changing 2sinxcosx = sin(2x), and my equation ended as
sin^2(2x) = sin(2x).

I subtracted sin(2x) by both sides and factored out sin(2x).
my equation ended like so:
sin^2(2x) - sin(2x) = sin(2x)*(sin(2x)-1)=0.

I was told to substitute 2x with some other variable to make it easier for me to solve this. So,
sin(2x)*(sin(2x)-1) = sin(y)*(sin(y)-1),
and siny = 0,1.
At this point, I am stuck because I don't know what to do when I change y back to 2x.
I would really appreciate your help with this!

p.s. Rest in peace, Mrs. Sue. She has saved my grades countless times and other students using this website as well.

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1. so, you have sin(2x) = 0 or sin(2x) = 1
You know that sin(0) = 0 and sin(π/2) = 1
So, 2x = 0 or 2x = π/2
Thus, x = 0 or x = π/4
The period of sin 2x is π, so the complete solution is

x = 0 + kπ
x = π/4 + kπ
for any integer k.

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2. Thank you so much for your help!

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