Solving trig equation: (sinx)^2=sinxcosx, on [0, 2pi)

q_fruit

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Jun 13, 2005
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Solve using exact values: (sinx)^2=sinxcosx, (can't be less than 0 and can't be bigger or equal to 2pi)

I've tried identities and what not but I don't know how to solve this w/o a calc.

T.i.a.
 
Re: Solving trig equation: sin2x = sinxcosx, on [0, 2pi)

q_fruit said:
I've tried identities and what not....
Which identities? What did you get?

I would assume you tried the double-angle formula, and came to an odd conclusion. But it should be noted that this conclusion depends upon your ability to divide through. Under what conditions can you not divide through?

Eliz.
 
Okay, I see. I can show you everything I've tried.

(sinx)^2 = (sin2x)/2
2((sinx)^2) - sin2x =0
and then I don't know how I can make it work.

1-(cosx)^2 = sinxcosx
1-(cosx)^2 -sinxcosx=0
1-cosx(cosx-sinx)=0
....and again not going anywhere.

I've tried some other things that just make no sense...so there yah go.
Thx.
 
q_fruit said:
o Dear! there was a mistake in the post!
it's (sinx)^2=sinxcosx sorry!
So, what's the problem? Do it.

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

sin(x) = 0 is pretty easy.

sin(x) - cos(x) = 0 is a little trickier

You can go this way: sin(x) = cos(x) That's pretty easy.
You can go this way: \(\displaystyle \sqrt{2}*sin(x-\frac{\pi}{4}) = 0\) Same deal.
You can go this way: sin(x) = cos(x) ==> tan(x) = 1
 
There was no problem. I just still didn't know how to do it.
Thank you for the help.
 
Yes I do actually understand.
I am studying for an exam, so my aim here is not to get a quicky answer. I know there are some people who ask questions here that just want a fast answer but I am not one of them. I don't want to start anything but sometimes when I try to ask a question on this forum I feel like the person who helps, is helping but giving me some sort of attitude at the same time because they assume I'm one of those people. I really appreciate the help but I just wanted to say that. I even consulted this question with a friend, and I guess the answer seems pretty easy to you guys but I'm in grade 12 and I'm not as bright as you all (obviously heh..).
 
Well, have a little more patience, with us and with yourself. All we have to go on is what you write. We don't have any way to KNOW who you are. Just answer our questions without assigning to them any value judgments. There is no reason to assume any "attitude".

The ONLY reason these problems look easy when I do them is that I have done several thousand in my lifetime, just like many other regulars here. There is no reason why you can't do the same, unless you give up. Practice, practice, and more practice.
 
I'm being completely patient. I have asked questions here before and I've gotten a "Are you really trying to tell me you can't do that?!" and then here you just said, "Do you know now or did I just do it for you?"
I mean that's kind of unnecessary.. do I have to "assume" the above quotes were attitude? They kind of just were.
Anyway this is a math forum. I apologize for turning this post into something else but I just had to say what was on my mind.
 
Sometimes, the questions asked are sufficiently fundamental that an inability to solve them may indicate that the student just isn't ready for the material. This would behoove us to help the student rethink much more than just the problem at hand.

For you, not knowing the difference between proving an identity and finding solutions to an equation is troublesome. It indicates that you are missing something important. I tried to phrase my statements in such a way as to give you an opportunity to indicate that you do know the difference, but you haven't responded well. In your case, I do not think it is a serious problem, just a little misunderstanding that can be cleared up. In any case, it requires a judgment call from me. That requires a little more patience from you.

We're glad you're here. Let's learn some math!
 
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