Let $$ f(x) = \begin{cases} \sin(x)&\text{if $x$ is rational}\\ 1-2\cos(x)&\text{otherwise} \end{cases} $$
We have to comment on the continuity of the function. Whether it is continuous at infinite points, one point or nowhere
My approach: I solved $\sin x=1-2\cos x$ and found that there were infinitely many solutions. But in this case where the function is defined differently for rational and irrational x does equality imply continuity?
Yes, the function
$$f(x)=\begin{cases}\sin(x)&\text{ if $x$ is rational,}\\ 1-2\cos(x) &\text{ if $x$ is rational.} \end{cases}$$ is continuous at $x$ if and only if $\sin(x)=1-2\cos(x)$. The set of solutions is infinite but discrete, so nothing to worry about.
The continuity at such points follows from the continuity of the functions $\sin(x)$ and $1-2\cos(x)$.
To show that $f$ is not continuous when the equation is not satisfied use the fact that for any real number $x$ there exists a sequence of irrationals $(y_n)_n$ and there exists a sequence of rationals $(z_n)_n$ such $y_n\to x$ and $z_n\to x$. Then $$\lim_{n\to\infty}f(z_n)=\lim_{n\to\infty}\sin(z_n)=\sin(x)\not=1-2\cos(x)=\lim_{n\to\infty}(1-2\cos(y_n))=\lim_{n\to\infty}f(y_n).$$