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(A) Given the parametric equations: $x = a(2 \cos t - \cos 2t)$ and $y = a(2 \sin t - \sin 2t)$.To find the slope of the tangent,we calculate $\frac{d

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$2\pi / 3$

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(A) Given the parametric equations: $x = a(2 \cos t - \cos 2t)$ and $y = a(2 \sin t - \sin 2t)$.To find the slope of the tangent,we calculate $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$.$\frac{dy}{dt} = a(2 \cos t - 2 \cos 2t) = 2a(\cos t - \cos 2t)$.$\frac{dx}{dt} = a(-2 \sin t + 2 \sin 2t) = 2a(\sin 2t - \sin t)$.Thus,$\frac{dy}{dx} = \frac{2a(\cos t - \cos 2t)}{2a(\sin 2t - \sin t)} = \frac{\cos t - \cos 2t}{\sin 2t - \sin t}$.For the tangent to be parallel to the $x$-axis,$\frac{dy}{dx} = 0$,which implies $\cos t = \cos 2t$.Using the identity $\cos A = \cos B \implies A = 2n\pi \pm B$,we have $2t = 2n\pi \pm t$.Case $1$: $2t = 2n\pi + t \implies t = 2n\pi$.Case $2$: $2t = 2n\pi - t \implies 3t = 2n\pi \implies t = \frac{2n\pi}{3}$.For $n=1$,$t = 2\pi/3$. For $n=2$,$t = 4\pi/3$.The difference between these values is $\frac{4\pi}{3} - \frac{2\pi}{3} = \frac{2\pi}{3}$.

At any two points of the curve represented parametrically by $x = a(2 \cos t - \cos 2t)$ and $y = a(2 \sin t - \sin 2t)$,the tangents are parallel to the $x$-axis. The values of the parameter $t$ corresponding to these points differ from each other by:

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