If cos(alpha - beta) = 1 and cos(alpha + beta | vedclass

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(D) Given $\cos(\alpha - \beta) = 1$ and $-\pi < \alpha, \beta < \pi$.Since $-\pi < \alpha < \pi$ and $-\pi < \beta < \pi$,we have $-2\pi < \alpha - \

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$4$

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(D) Given $\cos(\alpha - \beta) = 1$ and $-\pi Since $-\pi The equation $\cos(\alpha - \beta) = 1$ implies $\alpha - \beta = 0$,so $\alpha = \beta$.Substituting $\alpha = \beta$ into the second equation,we get $\cos(\alpha + \alpha) = \cos(2\alpha) = \frac{1}{e}$.Since $-\pi In the interval $(-2\pi, 2\pi)$,the equation $\cos(2\alpha) = \frac{1}{e}$ has solutions where $\cos(x) = \frac{1}{e}$ for $x \in (-2\pi, 2\pi)$.Since $0 Thus,there are $4$ distinct values for $\alpha$,and since $\alpha = \beta$,there are $4$ ordered pairs $(\alpha, \beta)$.

If $\cos(\alpha - \beta) = 1$ and $\cos(\alpha + \beta) = \frac{1}{e}$,where $-\pi < \alpha, \beta < \pi$,then the total number of ordered pairs $(\alpha, \beta)$ is:

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