Let $f:\left[0, 2\right]\to ℝ$ be the function defined by

$f\left(x\right)=\left(3-\sin \left(2\pi x\right)\right)\sin \left(\pi x-\frac{\pi }{4}\right)-\sin \left(3\pi x+\frac{\pi }{4}\right)$.

If $\alpha , \beta \in \left[0, 2\right]$ are such that $\left\{x\in \left[0, 2\right] :f\left(x\right)\ge 0\right\}=\left[\alpha , \beta \right]$, then the value of $\beta -\alpha$ is _____

Let $\pi x-\frac{\pi }{4}=\theta \Rightarrow \pi x=\theta +\frac{\pi }{4}$

$\because f\left(x\right)\ge 0\Rightarrow \left[3-\sin \left(2\theta +\frac{\pi }{2}\right)\right]\sin \theta -\sin \left(3\theta +\pi \right)\ge 0$

$\Rightarrow \left(3-\cos 2\theta \right)\sin \theta +\sin 3\theta \ge 0$

$\Rightarrow \left(2+2{\sin }^2\theta \right)\sin \theta +3\sin \theta -4{\sin }^3\theta \ge 0$

$\Rightarrow \sin \theta \left[5-2{\sin }^2\theta \right]\ge 0$

$\Rightarrow \sin \theta \ge 0\Rightarrow \theta \in \left[0, \pi \right]$

(Since, $x\in \left[0, 2\right]\Rightarrow \mathrm{\pi x}\in \left[0, 2\mathrm{\pi }\right]\Rightarrow \theta \in \left[-\frac{\pi }{4}, \frac{7\pi }{4}\right]$)

$\Rightarrow \sin \theta \ge 0\Rightarrow \theta \in \left[0, \pi \right]\Rightarrow \pi x\in \left[\frac{\pi }{4},\frac{5\pi }{4}\right]$

$\Rightarrow x\in \left[\frac{1}{4}, \frac{5}{4}\right]\Rightarrow \alpha =\frac{1}{4}, \beta =\frac{5}{4}$

Thus, $\beta -\alpha =\frac{5}{4}-\frac{1}{4}=1$.