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Is there a way to find a function that produces all the roots of a sine variant? I have the following function:

$5e^{-.2x} \left(-\frac{1}{5}\sin x + \cos x\right)$

Because $e^x$ is monotonic I only need to find roots of the second factor. I can find one root of it like this:

\begin{alignat*}{3} & & 0 ~=~ & -\frac{1}{5}\sin x + \cos x \\ & \implies & - \cos x ~=~ &-\frac{1}{5}\sin x \\ & \implies & \cos x ~=~ &\frac{1}{5}\sin x \\ & \implies & 5\cos x ~=~ &\sin x \\ & \implies & 5~=~ &\frac{\sin x}{\cos x} \\ & & ~=~ &\tan(x) \\ & \implies & \tan^{-1}(5) ~=~ &x \\ \end{alignat*}

But how can I infer a formula from this that gives me any other root?

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    $+ \pi ~ n , n \in \mathbb{Z}$. – 2017-01-23
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    @Moo I tried that already and I did not get $0$. You do mean $x_i = tan^{-1}(5 + \pi n)$, right? – 2017-01-23
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    Almost, you want $x_n = \tan^{-1}(5) + \pi ~ n, n \in \mathbb{Z}$. – 2017-01-23
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    @Moo Oh, ok, yea makes sense now. – 2017-01-23
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    It's not because $e^x$ is monotonic, but because it vanishes nowhere. – 2017-01-23

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