Hi there. In advance, I apologize if this question is too broad. Please do not downvote if that is the case, as this question is purely imaginative curiosity. I will close it should it turn out to be too broad.


So, this might be the most ridiculous inspiration ever. However, I have been reading Trigonometry for Dummies by Mary Jane Sterling. On pages 250 and 251, Sterling uses graphing to solve the following equations: $$ \begin{align} \cos (2x)&=2\cos(x).\\ \cos^2(x)-0.4\sin(x)&=0.6. \end{align} $$ It was clear to me that these two equations are just quadratics in terms of $\cos(x)$ and $\sin(x)$ respectively. This inspired me to implore as to why the author was motivated to use graphing to solve these two equations. I cannot see what motivates that, but I may very well be missing something. What follows is the true question that arose.


After considering the above, I asked myself: Are there messy trigonometric equations? By messy, I mean the sense that quintic equations are considered messy in general since they do not possess a general solution in terms of the basic arithmetical operations of addition, subtraction, extraction of roots, multiplication, and division. (Pardon me if I'm forgetting something here.)

I searched a little bit and there seems to be no resources on this. Here is what I am curious about: Is there a subfield of study akin to Galois theory (or perhaps an application of Galois theory?) that studies the solvability of trigonometric equations and the nature of the solutions to trigonometric equations? It should be noted that what I mean by 'trignometric equations' is any equation which uses any of the three trigonometric functions (sine, cosine, and tangent), addition, multiplication, subtraction, division, and extraction of roots.

P.S. As part of an answer to this question, I would like to see some pathological trigonometric equations if you are willing to provide them.

Further Thoughts

Looking at Wiki for the specifics, I see that . . .

. . . whether a polynomial was solvable or not was equivalent to whether or not the permutation group of its roots – in modern terms, its Galois group – had a certain structure – in modern terms, whether or not it was a solvable group.

In essence, if the Galois group of a polynomial is a solvable group, then the polynomial is solvable. Wiki goes on to explain . . .

This group was always solvable for polynomials of degree four or less, but not always so for polynomials of degree five and greater, which explains why there is no general solution in higher degree.

In this sense I am curious about the solutions of trigonometric equations. Is there any such concept similar to the Galois group that is applicable to the study of trigonometric equations as defined above? It should be noted I don't fully understand these concepts, and I may be completely off in my thinking. I apologize if that is the case.


The Weierstrass substitution $t=\tan(x/2)$ turns a rational function of trigonometric functions of $x$ into a rational function of $t$. Thus in principle the usual tools of Galois theory apply.

Alternately, the trigonometric functions of $x$ can be expressed as rational functions of $e^{ix}$, so again in principle we are in familiar territory.

This does not apply to "mixed" equations such as $x\sin x=1$. Also not apparently amenable to algebraic techniques are equations that, say, involve both $\sin x$ and $\sin(\alpha x)$, where $\alpha$ is irrational.

  • $\begingroup$ I don't understand the Weierstrass substitution completely. So, I apologize if this is an ignorant question: Are you saying that we can use the Weierstrass substitution to take any trigonometric equation and convert it to a state wherein the "usual tools of Galois theory apply"? Could you clarify this a bit, perhaps with an example? $\endgroup$ – 000 Nov 17 '12 at 0:34
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    $\begingroup$ $1+\cos^3 x=10\sin x$. Put $t=\tan(x/2)$. Then $\sin x=2t/(1+t^2)$, $\cos x=(1-t^2)/(1+t^2)$. Substitute, clear denominators. We obtain a polynomial equation in $t$. $\endgroup$ – André Nicolas Nov 17 '12 at 0:44
  • $\begingroup$ I get what you mean entirely. Thank you, André, your thought seems to be the most clever. $\endgroup$ – 000 Nov 17 '12 at 0:46
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    $\begingroup$ The substitutions I mentioned are widely useful. In particular, the use of complex exponentials is fundamental. $\endgroup$ – André Nicolas Nov 17 '12 at 0:54
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    $\begingroup$ An interesting instance of algebra mixing with equations involving trigonometric functions is the analysis of what angles are trisectable by straightedge and compass. $\endgroup$ – André Nicolas Nov 17 '12 at 0:59

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