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The Fundamental Theorem of Algebra states that "Any polynomial of degree $n$ ... has $n$ roots." Is there anything analogous for trigonometric equations? I've been solving some trigonometric equations, and solving some of the slightly more complex ones involves, what seems like guess and check, applying certain trig identities (namely $sin (\theta + 2\pi)$ or $sin(\pi - \theta) = sin (\theta)$) over and over again to check to see whether the resultants are within the given domain.

So, my question is, if we cant determine the exact solutions from a method other than guess and check, can we at least determine the number of solutions of a trigonometric equation given its range using a more mathematical procedure (i.e. no guess and check whatsoever)?

Widawensen
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StopReadingThisUsername
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  • After you have your equation in a form like $\sin x = a$ or what have you, I would suggest drawing a diagram of the function over one of its periods, with a line $y=a$ drawn in too. From that, you should get a good sense of how many solutions your function has per period, and then whatever solutions it has in one period, it will have images of those solutions over any other period of the function. Of course, getting in that form is what takes work. – user24142 Feb 02 '16 at 04:15

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Note $$ \sin\theta = \frac{e^{i\theta}-e^{-i\theta}}{2i},\qquad \cos\theta = \frac{e^{i\theta}+e^{-i\theta}}{2} $$ Any polynomial in $\sin\theta,\cos\theta$ is a rational function in $e^{i\theta}$. So solving such a thing is: solving a polynomial equation, taking logarithm, dividing by $i$. (Of course complex logarithm is multi-valued.)

GEdgar
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