# Solve trigonometric system of equations

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?

• 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 $0$ ? – Peter Aug 3 '17 at 16:15

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.

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 $$x-y = \frac{(2n+1)\pi}{2}, \; n \in \mathbb{Z}.$$

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