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Is there a way to solve for x in $\dfrac{\cos(ax)}{\cos(bx)} = c$?

This is similar to the question on $\dfrac{\cos^{-1}(ax)}{\cos^{-1}(bx)} = c$.

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To the best of my knowledge, there is no way to find a closed form solution for $x$ in general. – Jonathan Gleason Aug 22 '11 at 15:18
Do you know anything about the ratio a/b? – yatima2975 Aug 22 '11 at 15:19
@yatima2975: No. They represent more complicated expressions, but they're independent and arbitrary. – jnm2 Aug 22 '11 at 15:28
up vote 2 down vote accepted

The answer is similar to the other one.

If $a/b = m/n$ is rational then the equation can be written as $\cos (m\theta) = c \cos (n \theta)$ where $ax = m \theta$. For $c = \pm 1$, $\theta$ will be an easily described angle, and if $|c| \neq 1$, then $\cos(\theta)$ is an algebraic function of $c$ since the equation is equivalent to $P_m(\cos \theta) = P_n( \cos \theta)$ where the $P$ are Chebyshev polynomials. If $c$ is an algebraic number then so is $\cos(\theta)$.

For $a/b$ irrational and $c \neq \pm 1$ there is no algebraic solution.

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Your last statement is a bit confusing: it must be interpreted as "there is no algebraic method of solution", not "there is no algebraic number that is a solution". – Robert Israel Aug 22 '11 at 16:38
@Robert: yes, though actually I meant something stronger like "it is expected that the solution is not in a finite extension field of $Q(a,b,c)$, assuming as many standard conjectures as needed from transcendental number theory". – zyx Aug 24 '11 at 19:34

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