# Solve this equation

1. Feb 19, 2012

### kscplay

1. The problem statement, all variables and given/known data

Solve for x and y:
x2 + (√8)x*sin[(√2)xy] +2 = 0

2. Relevant equations

3. The attempt at a solution
Other than decomposing the root 8 i don't know what else to do. any hints? thanks.

2. Feb 19, 2012

### ehild

Hi kscplay,

Transform the equation into the form (x+a)2+b=0

ehild

3. Feb 20, 2012

### Staff: Mentor

4. Feb 20, 2012

### ehild

I meant (x+√2sin[√2xy])2=-2+2sin2[√2xy]).

The left hand side can not be negative. The right hand side ?...

ehild

5. Feb 20, 2012

### jkristia

this might be a silly question, but wouldn't you first isolate y?

6. Feb 20, 2012

### ehild

That could be also a way to find the solution. But I would look at the right hand side of my equation and see if it can be positive.

ehild

Last edited: Feb 20, 2012
7. Feb 20, 2012

### ehild

If we isolate y we get $$y=\frac{1}{\sqrt{2}x}\arcsin(-\frac{x^2+2}{2\sqrt{2}x})$$ The argument of arcsin has to be in the range [-1,1]. What does it mean for x?

My method of transforming the equation to $(x+\sqrt{2}\sin(\sqrt{2}xy)^2=-2+2sin^2(\sqrt{2}xy)$ shows that the left hand side can not be negative, the right-hand side can not be positive so both have to be zero.

$$x+\sqrt{2}\sin(\sqrt{2}xy)=0$$
$$\sin^2(\sqrt{2}xy)=1$$

two equations, two unknowns...

ehild

Last edited: Feb 21, 2012
8. Feb 21, 2012

### Mentallic

9. Feb 21, 2012

### ehild

There was a typo , I corrected it in post #7. The square should not be there in my first equation. Correctly it is $$x+\sqrt{2}\sin(\sqrt{2}xy)=0$$
ehild

Last edited: Feb 21, 2012
10. Feb 21, 2012

### kscplay

That's a smart idea. Thanks ehild :)

11. Feb 22, 2012

### ehild

You are welcome, kscplay.

The other way, jkristia suggested would work too.

$$y=-\frac{1}{\sqrt{2}x}\arcsin \left( \frac{x^2+2}{2\sqrt{2}x} \right)$$

The magnitude of the argument of the arcsin function can not exceed 1:
$$\left| \frac{x^2+2}{2 \sqrt{2} x} \right| \leq 1$$

The function

$$\frac{x^2+2}{2\sqrt{2}x}$$ has extrema at x=±√2, minimum (1) for positive x and maximum (-1) for negative x. Only these values are allowed for x.

ehild

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Last edited: Feb 22, 2012