Find the interval/intervals in which the function $\displaystyle \mathrm{f}(\mathrm{x})=\sin 3 \mathrm{x}-\cos 3 \mathrm{x}$, $\displaystyle 0<\mathrm{x}<\frac{\pi}{2}$ is strictly increasing.

$\displaystyle \mathbf{f}^{\prime}(\mathbf{x})=\mathbf{3} \boldsymbol{\operatorname { c o s }} \mathbf{3 x}+\mathbf{3} \boldsymbol{\operatorname { s i n }} \mathbf{3 x}$ \[\mathrm{f}^{\prime}(\mathrm{x})=0 \Rightarrow \sin 3 \mathrm{x}=-\cos 3 \mathrm{x} \Rightarrow \mathrm{x}=\frac{\pi}{4} \] For $\displaystyle \mathbf{x} \in\left(\mathbf{0}, \frac{\boldsymbol{\pi}}{\mathbf{4}}\right), \mathbf{3} \boldsymbol{\operatorname { c o s }} \mathbf{3} \mathbf{x}+\mathbf{3} \boldsymbol{\operatorname { s i n }} \mathbf{3} \mathbf{x} \boldsymbol{>} \mathbf{0}$ $\displaystyle \Rightarrow \mathbf{f}^{\prime}(\mathbf{x})>\mathbf{0}, \mathbf{f}$ is strictly increasing function in $\displaystyle \left(\mathbf{0}, \frac{\pi}{4}\right)$ or $\displaystyle \left(\mathbf{0}, \frac{\pi}{4}\right]$