Can anyone give me pointers for solving these simultaneous equations? $$\cos\pi x=0$$ $$\cos(\pi x)^2=0$$ There doesn't seem to be much information out there about how to work with $\cos(\pi x)^2$. I tried using the Taylor expansion $\cos x=\sum_{n=0}^∞ \frac{(-1)^n}{(2n)!}(x)^{2n}$, which gave me $$\cos\pi x=\sum_{n=0}^∞ \frac{(-1)^n}{(2n)!}(\pi x)^{2n}=0$$ $$\cos(\pi x)^2=\sum_{n=0}^∞ \frac{(-1)^n}{(2n)!}(\pi x)^{4n}=0$$ I divided by $\pi^2$ to get $$\sum_{n=0}^∞ \frac{(-1)^n}{(2n)!}x^{2n}=0$$ $$\sum_{n=0}^∞ \frac{(-1)^n}{(2n)!}x^{4n}=0$$ ...but then I got stuck. I made the (clearly wrong) assumption that $$\sum_{n=0}^∞ \frac{(-1)^n}{(2n)!}x^{2n}\equiv\left(\sum_{n=0}^∞ \frac{(-1)^n}{(2n)!}\right)\left(\sum_{n=0}^∞ x^{2n}\right)$$ This gave me $$\sum_{n=0}^∞ x^{2n}=0$$ $$\sum_{n=0}^∞ x^{4n}=0$$ Neither of these equations have real-number solutions.
Effectively, I have simultaneous equations in the form $$\sum_{n=0}^∞ f(x)g(x)=0$$ and I don't know how to proceed. Perhaps the whole Taylor series idea is a red herring?
I'd really appreciate your advice.
In the end, Taylor series aren't needed at all. $$\cos\pi x=0\iff x=n-\frac12\qquad n\in\mathbb Z$$ $$\cos(\pi x)^2=\cos\pi(\pi x^2)=0\iff\pi x^2=m-\frac12\qquad m\in\mathbb Z$$ $$\pi x^2=\pi\left(n-\frac12\right)^2=m-\frac12$$ $$\pi=\frac{m-\frac12}{\left(n-\frac12\right)^2}$$ But $\pi$ is irrational and the denominator is never zero, so there are no solutions to the simultaneous equations.