Showing that $\sin(\sqrt{4 \pi^{2}n^{2} + x})$ converges uniformly on $[0,1]$

Suppose we are considering the sequence of functions $f_{n}(x)=\sin(\sqrt{4 \pi^{2}n^{2} + x})$ and I am having trouble showing that that $f_{n}$ converges uniformly on the interval $[0,1]$.

An idea, I've tried is to consider the Taylor series:

$$\sin(\sqrt{4 \pi^{2}n^{2} + x}) = (\sqrt{4 \pi^{2}n^{2} + x})- \frac{(\sqrt{4 \pi^{2}n^{2} + x})^{3}}{6} + O((\sqrt{4 \pi^{2}n^{2} + x})^{5})$$

but I haven't gotten anything useful as of yet.

Fix $x\in[0,1]$. The Mean Value Theorem shows that $$|\sqrt{4\pi^2n^2+x}-2\pi n|<\frac x{4\pi n}\le\frac1{4\pi n}.$$ So $2\pi n \le \sqrt{4\pi^2n^2+x} < 2\pi n +1/(4\pi n)$. Then apply MVT again to show$$|\sin(\sqrt{4\pi^2n^2+x})| =|\sin(\sqrt{4\pi^2n^2+x})-\sin(2\pi n)|<\frac1{4\pi n}.$$

Your sequence converges uniformly to $0$.

\begin{align} \left|\sin\left(\sqrt{4\pi^2n^2+x}\right)\right| &= \left|\sin\left(\sqrt{4\pi^2n^2+x}\right)-\sin(2\pi n)\right| \\ &= \left|2\sin\left(\frac{\sqrt{4\pi^2n^2+x} - 2\pi n}2\right)\cos\left(\frac{\sqrt{4\pi^2n^2+x} + 2\pi n}2\right)\right|\\ &\le 2\left|\sin\left(\frac{\sqrt{4\pi^2n^2+x} - 2\pi n}2\right)\right|\\ &\le 2\left|\frac{\sqrt{4\pi^2n^2+x} - 2\pi n}2\right|\\ &= \sqrt{4\pi^2n^2+x} - 2\pi n\\ &= \frac{4\pi^2n^2+x - 4\pi^2n^2}{\sqrt{4\pi^2n^2+x} + 2\pi n}\\ &= \frac{x}{\sqrt{4\pi^2n^2+x} + 2\pi n}\\ &\le \frac{1}{4\pi n} \xrightarrow{n\to\infty} 0 \end{align}

uniformly in $x \in [0,1]$.