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

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(D) Given $\cos(\alpha - \beta) = 1$ and $\cos(\alpha + \beta) = 1/e$,where $\alpha, \beta \in [-\pi, \pi]$.From $\cos(\alpha - \beta) = 1$,we have $\

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$4$

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(D) Given $\cos(\alpha - \beta) = 1$ and $\cos(\alpha + \beta) = 1/e$,where $\alpha, \beta \in [-\pi, \pi]$.From $\cos(\alpha - \beta) = 1$,we have $\alpha - \beta = 0$,which implies $\alpha = \beta$.Substituting $\alpha = \beta$ into the second equation,we get $\cos(2\alpha) = 1/e$.Since $\alpha, \beta \in [-\pi, \pi]$,the range of $2\alpha$ is $[-2\pi, 2\pi]$.We know that $0 Since $\alpha = \beta$,each value of $\alpha$ gives a unique pair $(\alpha, \beta)$.Therefore,there are $4$ such pairs.

If $\cos(\alpha - \beta) = 1$ and $\cos(\alpha + \beta) = 1/e$,where $\alpha, \beta \in [-\pi, \pi]$,the number of pairs of $(\alpha, \beta)$ which satisfy both equations is:

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