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General Solutions

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Reply 20
Avatar for f1mad
f1mad
13
Original post by Lord of the Flies
Haha, I'm going to have to say the same (again!). I think I was unclear.


:tongue:.

Re. sin2x = sinx

2x = x+ 2npi

or

2x= 2npi+pi -x

-> x= 2npi or 3x= 2npi+pi
-> x = 2npi, 2npi/3 + pi/3

I don't really see the issue here though; the OP asked for a "general solution". And the aforementioned is the right way to go about it.
Original post by f1mad
:tongue:.

Re. sin2x = sinx

2x = x+ 2npi

or

2x= 2npi+pi -x

-> x= 2npi or 3x= 2npi+pi
-> x = 2npi, 2npi/3 + pi/3

I don't really see the issue here though; the OP asked for a "general solution". And the aforementioned is the right way to go about it.


but you lose the cos solutions
Reply 22
Avatar for f1mad
f1mad
13
Original post by Lord of the Flies
That wasn't really the problem, since I mentioned periodicity in my post. As Mr M pointed out:

sin⁡2x=sin⁡x⇒2x=x+nπ\sin 2x=\sin x\Rightarrow 2x=x+n\pi does not give the full solution.


Except, that is the incorrect GS.

See above :biggrin:.
Original post by f1mad
Except, that is the incorrect GS.

See above :biggrin:.


That's what I said. I don't understand what the fuss is about. :s-smilie: My post says "the same thought process does not give the full solution with sin". (which is the same as saying it is "the incorrect GS").
Reply 24
Avatar for f1mad
f1mad
13
Original post by TenOfThem
but you lose the cos solutions


I can't see where?

2sinxcosx = sinx

Given sinx =! 0

2cosx =1 -> cosx = 1/2 -> x= pi/3, 5pi/3 etc

Considering sinx = 0 -> x= 0, pi, 2pi etc

Is that not the same output as my GS? From where I'm sat it does.
Reply 25
Avatar for Cephalus
Cephalus
This whole thread has now confused me
Reply 26
Avatar for f1mad
f1mad
13
Original post by Lord of the Flies
That's what I said. I don't understand what the fuss is about. :s-smilie: My post says "the same thought process does not give the full solution with sin". (which is the same as saying it is "the incorrect GS").


Yeah, but, sin, cos and tan work differently when dealing with GS's anyway (which the OP should be aware about?). A minor thing to point out, but yeah :lol:.
Reply 27
Avatar for Ganhad
Ganhad
OP
2
i still dont get it -_-
Original post by Ganhad
i still dont get it -_-


You have been given two approaches.

Simply write down the general solution (if you understand it).

Or do this:

tan⁡2x=tan⁡x\tan 2x = \tan x

2tan⁡x1−tan⁡2x=tan⁡x\frac{2 \tan x}{1-\tan^2 x}=\tan x

2tan⁡x=tan⁡x(1−tan⁡2x)2 \tan x = \tan x (1-\tan^2 x)

Form a cubic and solve.
Reply 29
Avatar for Ganhad
Ganhad
OP
2
the general solution is n pi +alpha how would i do it with that
(edited 11 years ago)
Reply 30
Avatar for atsruser
atsruser
11
Original post by Ganhad
i still dont get it -_-

If you have, say, tan⁡x=tan⁡π4\tan x = \tan \frac{\pi}{4}, then you can see a solution immediately; it's x=π4x = \frac{\pi}{4}. This should be obvious.

However, you probably know that the tantan function is periodic, and its graph comprises an infinite set of copies of the same curve, each offset by π\pi. So although x=π4x = \frac{\pi}{4} is a solution to the original equation, anothe solution is x=π4+πx = \frac{\pi}{4} + \pi, as is x=π4−πx = \frac{\pi}{4} - \pi as is x=π4+2πx = \frac{\pi}{4} + 2\pi and so forth. (Look at the graph of tan⁡x\tan x to convince yourself of this fact).

In fact any number of the form x=π4+nπx = \frac{\pi}{4} + n\pi where n∈Zn \in \mathbb{Z} is a solution of the original equation.

Now what if we have the equation tan⁡x=tan⁡y\tan x = \tan y ? Well, *exactly* the same line of reasoning works. We can immediately write down the general solution of the equation as x=y+nπx = y + n\pi.

And this works even if y=2xy = 2 x. So the general solution to your equation is given, by this argument, as x=2x+nπx = 2x + n\pi.

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