Figuring out when $f(x)=\mathrm{sin}({x}^{2})$ is increasing and decreasing

Regarding the function $f(x)=\mathrm{sin}({x}^{2})$, I'm supposed to figure out when it is increasing/decreasing.

So far, I've found the derivative to be ${f}^{\prime}(x)=2x\mathrm{cos}({x}^{2})$.

So long as I can solve the inequality $\mathrm{cos}({x}^{2})>0$. I can figure the rest out, but this is where I'm stuck.

I've narrowed it down to $-\frac{\pi}{2}<{x}^{2}<\frac{\pi}{2}$ meaning ${x}^{2}<\frac{\pi}{2}\pm n2\pi ,\text{}\text{}n\in \mathbb{N}$.

Then I get $x<\pm \sqrt{\frac{\pi}{2}\pm n2\pi}$.

Am I on the right track here? I can't seem to find a path from here.

Regarding the function $f(x)=\mathrm{sin}({x}^{2})$, I'm supposed to figure out when it is increasing/decreasing.

So far, I've found the derivative to be ${f}^{\prime}(x)=2x\mathrm{cos}({x}^{2})$.

So long as I can solve the inequality $\mathrm{cos}({x}^{2})>0$. I can figure the rest out, but this is where I'm stuck.

I've narrowed it down to $-\frac{\pi}{2}<{x}^{2}<\frac{\pi}{2}$ meaning ${x}^{2}<\frac{\pi}{2}\pm n2\pi ,\text{}\text{}n\in \mathbb{N}$.

Then I get $x<\pm \sqrt{\frac{\pi}{2}\pm n2\pi}$.

Am I on the right track here? I can't seem to find a path from here.