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I'm trying to solve this system:

$$\begin{align} 11\cos x_1 + 12\cos x_2 &= \phantom{1}0\\ 11\sin x_1 + 12\sin x_2 &= 17 \end{align}$$

I think there is no answer, but I'm not sure. If you could show me a proof (or show me that I'm wrong, and there is a value for $x_1$ and $x_2$ that satisfies the equations), that would be great! Thanks!

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Hint:

$$11\cos(x_1) + 12\cos(x_2) = 0\implies 11\cos(x_1)=-12\cos(x_2)$$
$$11\sin(x_1) + 12\sin(x_2) = 17 \implies 11\sin(x_1)=17-12\sin(x_2)$$

Now square both equations and add them then replace $\cos^2(x_i)+\sin^2(x_i)=1$.

$$11^2(\cos^2(x_1)+\sin^2(x_1))=12^2\cos^2(x_2)+17^2-2\cdot 17 \cdot 12 \sin(x_2)+12^2\sin^2(x_2)$$ $$11^2=12^2+17^2-2\cdot 17 \cdot 12\sin(x_2)$$

Can you solve it from here?

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From these one gets $11e^{ix_1}+12e^{ix_2}=17i$. So the points $0$, $11e^{ix_1}$ and $17i$ form a triangle in the Argand diagram with side-lengths $11$, $12$ and $17$. One can find the possible values of $11e^{ix_1}$ by intersecting the circle centre $0$ and radius $11$ with the circle centre $17i$ and radius $12$.

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