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P.S. \[0 \le x < 2\]

Use identity of 2sinxcosx = sin2x

I tried, but then I got sin 2x = 1/2, so what?

I mean sin2x = 1

First, you would get sin2x =1, not 1/2. Agreed?

Yes

Ok, so you know where sin m = 1, and that's at pi/2 + 2kpi. Good so far?

Okay but keep in mind that \[0 \le x < 2\pi\]

So, those are the only 2 values for x.

Hmmm, let me see if I get this

Go ahead, ask questions. I'll try to clarify, but it looks like you already understand most of this.

I don't understand why x = (5/4)pi is also a solution

3/2*pi should also be a value for m, right?

Hmmm nevermind my last question

And how did you find out 5/4pi is an answer? You just guessed?

Hmmm but actually the constraint is 0<= x < 2pi

The "restriction" for 2x is that 2x has to be less than 4pi, not 2pi.

This problem is like my avatar. See how it is cyclic and goes offshing point? Very Zen!

that is off to a vanishing point.

You're a very good student, I can tell!

Math is way cool. Intuitive and analytical at the same time!

That's true. I'm always thinking of how math is all about creativity.

I'll definitely look it up! Thanks!

You'll love it I guarantee it!

Watch the movie first if you don't have a lot of time, though. Actually a great movie.

So gotta go, but wonderful working with you and thx for the recognition medal!

Goodbye! Thanks and keep in touch!

Will do!