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(B) Let $	heta = \pi x - \frac{\pi}{4}$. Since $x \in [0, 2]$,$	heta \in [-\frac{\pi}{4}, \frac{7\pi}{4}]$.Then $2\pi x = 2	heta + \frac{\pi}{2}$,s

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(B) Let $	heta = \pi x - \frac{\pi}{4}$. Since $x \in [0, 2]$,$	heta \in [-\frac{\pi}{4}, \frac{7\pi}{4}]$.Then $2\pi x = 2	heta + \frac{\pi}{2}$,so $\sin(2\pi x) = \cos(2	heta)$.Also $3\pi x + \frac{\pi}{4} = 3(	heta + \frac{\pi}{4}) + \frac{\pi}{4} = 3	heta + \pi$,so $\sin(3\pi x + \frac{\pi}{4}) = \sin(3	heta + \pi) = -\sin(3	heta)$.The inequality $f(x) \geq 0$ becomes $(3 - \cos(2	heta)) \sin 	heta - (-\sin(3	heta)) \geq 0$.$(3 - (1 - 2\sin^2 	heta)) \sin 	heta + \sin(3	heta) \geq 0$.$(2 + 2\sin^2 	heta) \sin 	heta + (3\sin 	heta - 4\sin^3 	heta) \geq 0$.$2\sin 	heta + 2\sin^3 	heta + 3\sin 	heta - 4\sin^3 	heta \geq 0$.$5\sin 	heta - 2\sin^3 	heta \geq 0 \Rightarrow \sin 	heta (5 - 2\sin^2 	heta) \geq 0$.Since $5 - 2\sin^2 	heta > 0$ for all $	heta$,we must have $\sin 	heta \geq 0$.Thus,$	heta \in [0, \pi]$.$0 \leq \pi x - \frac{\pi}{4} \leq \pi$ $\Rightarrow \frac{\pi}{4} \leq \pi x \leq \frac{5\pi}{4}$ $\Rightarrow x \in [\frac{1}{4}, \frac{5}{4}]$.Therefore,$\alpha = \frac{1}{4}$ and $\beta = \frac{5}{4}$,so $\beta - \alpha = 1$.

Let $f:[0,2] \rightarrow R$ be the function defined by $f(x)=(3-\sin(2\pi x)) \sin(\pi x-\frac{\pi}{4})-\sin(3\pi x+\frac{\pi}{4})$. If $\alpha, \beta \in[0,2]$ are such that $\{x \in[0,2]: f(x) \geq 0\}=[\alpha, \beta]$,then the value of $\beta-\alpha$ is:

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