Solving e^x and y for x & y: Why is it Wrong?

[theBEAST]

Homework Statement

e^x*cosy=0
-e^x*siny=0

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?

 

[tiny-tim]

hi theBEAST!

(try using the X² button just above the Reply box )

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?

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!​

 

[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.

 

[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 $±1 ≠ 0.$