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I have a seemingly simple system of equations, but I don’t know how to solve this.

$$ \left\{ \begin{array}{ll} \cos(x) \cos(y) &=0 \\ - \sin(x) \sin(y) &=0 \\ \end{array} \right. $$

Is there any trick?

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    $\begingroup$ Hint : $\cos(x)$ and $\sin(x)$ cannot be both $0$, same for $\cos(y)$ and $\sin(y)$. So which combinations remain to get both expressions $0$ ? $\endgroup$ – Peter Aug 3 '17 at 16:15
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If $\cos{x}=0$ then since, $\sin{x}\neq0$, we get $\sin{y}=0$ and $$\left\{\left(\frac{\pi}{2}+\pi k,\pi m\right)|\{k,m\}\subset\mathbb Z\right\}.$$

The case $\cos{y}=0$ for you.

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Notice that we could set the equations equal $$ \cos(x)\cos(y) + \sin(x)\sin(y) = 0,$$ and using a trig identity, we get $$ \cos(x-y) = 0.$$ Then we know where the zeros of the cosine function are, namely when $$x-y = \frac{(2n+1)\pi}{2}, \; n \in \mathbb{Z}.$$

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  • $\begingroup$ And from there how do I get separate solutions for x and y? $\endgroup$ – philmcole Aug 24 '17 at 15:05

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