Solve f(t)=sin^2t|(t)

$$f ( t ) = \sin ^ { 2 } t | ( t )$$

Solve for f

$$\left\{\begin{matrix}f=\frac{|t|\left(\sin(t)\right)^{2}}{t}\text{, }&t\neq 0\\f\in \mathrm{R}\text{, }&t=0\end{matrix}\right.$$

Solve for t

$$\left\{\begin{matrix}\\t=0\text{, }&\text{unconditionally}\\t=-\arcsin(\sqrt{f})+2\pi n_{1}\text{, }n_{1}\in \mathrm{Z}\text{, }n_{1}\geq \frac{\arcsin(\sqrt{f})}{2\pi }\text{; }t=\arcsin(\sqrt{f})+2\pi n_{2}+\pi \text{, }n_{2}\in \mathrm{Z}\text{, }n_{2}\geq -\frac{\arcsin(\sqrt{f})+\pi }{2\pi }\text{; }t=\arcsin(\sqrt{f})+2\pi n_{3}\text{, }n_{3}\in \mathrm{Z}\text{, }n_{3}\geq -\frac{\arcsin(\sqrt{f})}{2\pi }\text{; }t=-\arcsin(\sqrt{f})+2\pi n_{4}+\pi \text{, }n_{4}\in \mathrm{Z}\text{, }n_{4}\geq -\frac{-\arcsin(\sqrt{f})+\pi }{2\pi }\text{, }&f\geq 0\text{ and }f\leq 1\\t=-\arcsin(\sqrt{-f})+2\pi n_{5}\text{, }n_{5}\in \mathrm{Z}\text{, }n_{5}\leq \frac{\arcsin(\sqrt{-f})}{2\pi }\text{; }t=\arcsin(\sqrt{-f})+2\pi n_{6}+\pi \text{, }n_{6}\in \mathrm{Z}\text{, }n_{6}\leq -\frac{\arcsin(\sqrt{-f})+\pi }{2\pi }\text{; }t=\arcsin(\sqrt{-f})+2\pi n_{7}\text{, }n_{7}\in \mathrm{Z}\text{, }n_{7}\leq -\frac{\arcsin(\sqrt{-f})}{2\pi }\text{; }t=-\arcsin(\sqrt{-f})+2\pi n_{8}+\pi \text{, }n_{8}\in \mathrm{Z}\text{, }n_{8}\leq -\frac{-\arcsin(\sqrt{-f})+\pi }{2\pi }\text{, }&f\geq -1\text{ and }f\leq 0\end{matrix}\right.$$