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#1 2007-10-21 07:07:21

tony123
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Registered: 2007-08-03
Posts: 228

Find real solutions

Find real solutions of the system

sin x + 2 sin(x+y+z) = 0,
sin y + 3 sin(x+y+z) = 0,
sin z + 4 sin(x+y+z) = 0.

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#2 2007-10-21 07:45:45

JaneFairfax
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Registered: 2007-02-23
Posts: 6,868

Re: Find real solutions

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#3 2007-10-22 21:46:53

tony123
Member
Registered: 2007-08-03
Posts: 228

Re: Find real solutions

but how JaneFairfax

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#4 2007-10-22 23:03:40

JaneFairfax
Member
Registered: 2007-02-23
Posts: 6,868

Re: Find real solutions

So you have

The solution to this is sinx = siny = sinz = 0.

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#5 2007-10-25 05:46:52

tony123
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Registered: 2007-08-03
Posts: 228

Re: Find real solutions

PLESE step by step

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#6 2007-10-25 06:30:40

TheDude
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Registered: 2007-10-23
Posts: 361

Re: Find real solutions

Take each individual equation and solve for sin(x + y + z).

So, from your first equation we get 2sin(x + y + z) = -sin(x) ---> sin(x + y + z) = -(1/2)(sin(x)).  Doing the same for the other 2 equations gives you sin(x + y + z) = -(1/3)(sin(y)) and sin(x + y + z) = -(1/4)(sin(z)).

Now, since sin(x + y + z) appears by itself in all 3 equations, we know the right side of each equation is equal to each other, so we now have -(1/2)(sin(x)) = -(1/3)(sin(y)) = -(1/4)(sin(z)).  Multiply through by -12 to get 6sin(x) = 4sin(y) = 3sin(z).

Take the first equation, 6sin(x) = 4sin(y).  Since both sides are of the form a*sin(w), where a is a constant and w is a variable, the graphs of both with have the same period and initial position.  The only difference is the amplitude, which will cause the graphs not to touch anywhere except where sin(w) = 0.  This same argument can be used for 3sin(z).

This gives us our final equation of sin(x) = sin(y) = sin(z) = 0.  As you probably know, this is true when x, y, and z are any integers multiplied by pi.


Wrap it in bacon

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#7 2007-11-01 03:19:08

JaneFairfax
Member
Registered: 2007-02-23
Posts: 6,868

Re: Find real solutions

TheDude wrote:

Take the first equation, 6sin(x) = 4sin(y).  Since both sides are of the form a*sin(w), where a is a constant and w is a variable, the graphs of both with have the same period and initial position.  The only difference is the amplitude, which will cause the graphs not to touch anywhere except where sin(w) = 0.

No, this is not correct. They will all have the same period, but they may not have the same initial position – in which case they may intersect at places other than where sin(w) = 0. For example, if we consider just the equation 6sin(x) = 4sin(y) alone, we see that it can be satisfied by x = π⁄6, y = sin[sup]−1[/sup](0.75).

In fact, I’ve just realized that I was rather hasty too to reach my conclusion in my working. sad

Last edited by JaneFairfax (2007-11-01 03:26:09)

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#8 2007-11-01 05:31:12

TheDude
Member
Registered: 2007-10-23
Posts: 361

Re: Find real solutions

I've been giving out a lot of wrong answers.

Going back to our equation 6sin(x) = 4sin(y) = 3sin(z), solve for both sin(x) and sin(y) in terms of z:
sin(x) = (1/2) sin(z)
sin(y) = (3/4) sin(z)

Now take the arcsin of these equations:
x = arcsin( (1/2) sin(z) )
y = arcsin( (3/4) sin(z) )

We know this operation is legal since the range of the sin function is -1 to 1, which is the domain of the arcsin function.  By multiplying sin(z) by a value less than 1 we stay within arcsin's domain.

From here choose any real value for z and plug it into these equations for a legal triple.  Stated more mathematically:


Note that there's probably a way of simplifying from here, but I don't know how.


Edit: Well, on second thought it doesn't seem to work out very well.  Taking Jane's example of x = pi/6, y = arcsin(3/4), and z = pi/2 we get sin(x) = 1/2, sin(y) = 3/4, and sin(z) = 1.  This means that we need sin(x + y + z) = -1/4 to solve the original 3 equations.  Unfortunately, sin(x + y + z) in this example equals 0.1978.....  It appears that something is wrong here.

Last edited by TheDude (2007-11-02 02:40:56)


Wrap it in bacon

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#9 2007-11-02 07:45:06

tony123
Member
Registered: 2007-08-03
Posts: 228

Re: Find real solutions

We have

.
Adding the first two equations and substracting the third, we have:



  (1)

If


Then

In the same way if one of the others two factors from (1) is 0

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