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?