Let alpha be the period of 3 sin frac{pi x}{3 | vedclass

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(A) The period of $f(x) = 3 \sin \frac{\pi x}{3} - \cos \frac{\pi x}{2} + 	an \frac{\pi x}{4}$ is the $LCM$ of the periods of its components. The per

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$\frac{3}{2}$

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(A) The period of $f(x) = 3 \sin \frac{\pi x}{3} - \cos \frac{\pi x}{2} + 	an \frac{\pi x}{4}$ is the $LCM$ of the periods of its components. The periods are $T_1 = \frac{2\pi}{\pi/3} = 6$,$T_2 = \frac{2\pi}{\pi/2} = 4$,and $T_3 = \frac{\pi}{\pi/4} = 4$. Thus,$\alpha = 	ext{LCM}(6, 4, 4) = 12$. For $\beta$,use the identity $\sin^2 A - \sin^2 B = \sin(A+B)\sin(A-B)$. Here,$\sin^2 \left( \frac{\pi}{7} + \frac{x}{4} ight) - \sin^2 \left( \frac{\pi}{7} - \frac{x}{4} ight) = \sin \left( \frac{2\pi}{7} ight) \sin \left( \frac{x}{2} ight)$. The period is $\beta = \frac{2\pi}{1/2} = 4\pi$. For $\gamma$,$\cos^4 x + \sin^4 x = 1 - 2 \sin^2 x \cos^2 x = 1 - \frac{1}{2} \sin^2(2x) = 1 - \frac{1}{2} \left( \frac{1 - \cos(4x)}{2} ight) = \frac{3}{4} + \frac{1}{4} \cos(4x)$. The period is $\gamma = \frac{2\pi}{4} = \frac{\pi}{2}$. Finally,$\frac{\alpha \gamma}{\beta} = \frac{12 	imes \frac{\pi}{2}}{4\pi} = \frac{6\pi}{4\pi} = \frac{3}{2}$.

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