Can somebody help me solve this one?
sinx*cos³x + sin³x*cosx = 1/2
I've got:
cos²x(sinx*cosx) + sin²x(sinx*cosx) = 1/2
(sin²x + cos²x)(sinx*cosx) = 1/2
Since sin²x + cos²x = 1
sinx*cosx = 1/2
How should I proceed now?

- anonymous

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- schrodinger

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- anonymous

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

- anonymous

Use identity of 2sinxcosx = sin2x

- anonymous

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

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## More answers

- anonymous

I mean sin2x = 1

- anonymous

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

- anonymous

Yes

- anonymous

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

- anonymous

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

- anonymous

np. Your FIRST value is pi/4 because sin m = sin 2x where m = pi/2 so x = pi/4 You SECOND and only other value is at x = (5/4)pi because when that is doubled, you are right back at pi/2

- anonymous

So, those are the only 2 values for x.

- anonymous

Hmmm, let me see if I get this

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

Hmmm nevermind my last question

- anonymous

np. After I show you, I think you'll realize that you already knew the answer, which is good, because you'll get the answers on your own then. If you take (5/4)pi as "x" and multiply by 2, you will get (5/2)pi, which is 2pi + pi/2, so you "cycled" back to pi/2 by going around the circle once and then went pi/2.
(3/2)pi will not be an answer because doubled it's 3pi which is 180 degrees and you have sin 180 = 0.

- anonymous

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

- anonymous

The trick is why did I work "backwards" and select (5/4)pi and then double it to "magically" arrive at pi/2, going once around the circle. If that's how I worked it, THAT would be confusing, so I'll explain how I did it. Sometimes you can work "backwards" to get an intuitive idea of where to go, but that doesn't give you the "how to get the answer". You can be intuitive and work backward to "see what you want" and what is really there at the same time. You then have to work forward, the logical way, so that you can "prove" what needs to be proved. So here goes...

- anonymous

We know that we have the constraint of 0 <= x <= 2pi. Good. However, our final equation is sin2x = 1. The left side 2x tells us that as x goes from 0 to 2pi, then 2x goes from 0 to 4 pi. Lightbulb time! We go around the circle TWICE! That's the key! So, before resloving x and 2x, we know we have sin (something) = 1 at pi/2 and 5pi/2 because we supposed to twice around. Again, that's the key. Twice around. So, look for halves of those values because x is doubled. So you start intuitively, but back it all up logically. Ask more questions.

- anonymous

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

- anonymous

That's fine. You'll notice that x = pi/4 and x = 5pi/4 and both of those ARE less than 2pi. It doesn't matter that 2x = 5pi/2 > 2pi. The restriction is on x which we are working with. The restriction is not on 2x.

- anonymous

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

- anonymous

Believe me, I can understand how this could be a little disturbing at first. Keep going over it, and you will see it. You are not expected to see it immediately, but I'm positive about this. If it helps, think of 2x as another variable of think of 2x as a f(x). Or think "twice around" for 2x is "once around" for x.

- anonymous

Hmmm I think I get it. But how would I go about writing that? I left at sin 2x = 1. So let 2x = m and we have m = pi/2. Therefore, 2x = pi/2 and x = pi/4.
What about explaining how I got 5/4pi and 5/2pi? I'd need to prove that in an exam, right?

- anonymous

The way I'd think about it if I put myself in the position of seeing it for the first time would be to say 1) where is sin m = 1? It's at pi/2 and all 2kpi additions to that. 2) I'd know to concentrate on pi/2 right off, but what about those other potential values? Are they "within range". 3) You'd have to relate 0 <= x <= 2pi and 0 <= 2x <= 4pi because that is like a function with y = 2x. You get to use all legitimate values of "Y". And you HAVE to or else you'll get only half credit for your answer. Think of how the sine function is periodic and repeats. That's your conceptual ace in the hole.

- anonymous

Another help: your first value was sin pi/2 = 1, but that's sin 2(pi/4) = 1 for your "x" value because you are working with 2x, so that is your intuitive inclination to consider other values. You had to double something a lot smaller.

- anonymous

You have to mull things like this over. But don't worry, they expand your mind, they really do. Just be comfortable with it and then it flows.

- anonymous

Wow, you're a great teacher. I get it now, thanks! I always have a little trouble considering values that aren't so apparent at first.

- anonymous

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

- anonymous

that is off to a vanishing point.

- anonymous

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

- anonymous

Thanks! The function part is very interesting, I didn't know it was okay to think of the restrictions like that!

- anonymous

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

- anonymous

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

- anonymous

That's it! There is an absolutely great biography about John Nash and his math exploits. It was made into a good movie called "A Beautiful Mind". The written account is way better though. All about the creative thinking process.

- anonymous

I'll definitely look it up! Thanks!

- anonymous

You'll love it I guarantee it!

- anonymous

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

- anonymous

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

- anonymous

Goodbye! Thanks and keep in touch!

- anonymous

Will do!

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