# Solving for x and y

1. May 24, 2012

### theBEAST

1. The problem statement, all variables and given/known data
e^x*cosy=0
-e^x*siny=0

3. The attempt at a solution
Since they both equal to 0 I set them equal to each other:
e^x*cosy = -e^x*siny
I can cancel the e^x and I get:
tany = -1

Thus y = 7π/4 + n where n is an integer.

However this is incorrect when I plug it back into the original set of equations. Why is this wrong?

2. May 24, 2012

### tiny-tim

hi theBEAST!

(try using the X2 button just above the Reply box )
because you threw away information

you started with two equations, and you ended with only one

you need two independent equations (for example, by subtracting instead of adding)

alternatively, just solve each original equation separately!!

3. May 24, 2012

### Infinitum

What you get is correct, but you missed a further step which disproves the existence of a solution. Put $y=3\pi/4$ in $cos(y)$ for the first equation, and you get

$e^xcos(y) = 0$

$-e^x = 0$

This tells you that there can be no real value of x so that this equation holds. Hence the equations have no solution.

4. May 24, 2012

### middleCmusic

As posted above, this system has no solution for real x and y. Here's a slightly different line of reasoning.

You started off with

$$e^x cos(y)=0$$
$$-e^x sin(y)=0$$

Note that in the second equation, we can get rid of the minus sign:

$$e^x cos(y)=0$$
$$e^x sin(y)=0$$

Now, we know that $e^x$ is never equal to $0$ for all $x \in ℝ$, so in each equation, we must have the other term in the product equal to $0$. That is, we must have $cos(y) = 0 = sin(y)$.

But from our knowledge of the unit circle, we know that this is never possible. In fact, since $sin^2(x)+cos^2(x)=1$, whenever either function is equal to $0$, the other is equal to $±1 ≠ 0.$