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I have the following general equation in $x$

$$a\cos(b - cx) - d\cos(e - fx) = 0$$

with constants $a,b,c,d,e,f$.

Is there an algerbraic solution to this or only a numeric one?

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The first $e$ is meant to be $d$? – Gerry Myerson Jan 16 '13 at 10:35
Yeah sorry! Fixed – bradgonesurfing Jan 16 '13 at 11:10

In the case $a=d$, the equation can be factored as $a(\cos(b-cx)-\cos(e-fx))=0$. If $a=0$, then $x$ can be anything. If $a\neq0$, then you must have $$\cos(b-cx)=\cos(e-fx).$$ This implies that the two arguments must be equal, or one the negative of the other, mod $2\pi$. Thus, $$ b-cx=e-fx + 2\pi k \textbf{ or }-(b-cx)=e-fx+2\pi k, $$ where $k\in\mathbb{Z}$.

The first possibility gives $$x=\frac{e-b+2\pi k}{f-c}.$$ This exhibits a single solution when $c\neq f$. If $c=f$, then we need $e-b$ to be a multiple of $2\pi$ for any solution to exist. In that case, the two curves are the same, potentially phase shifted, and every value of $x$ is a solution.

The second possibility gives $$x=\frac{e+b+2\pi k}{f+c}.$$ Similarly, there is only one solution when $f+c\neq 0\Rightarrow f\neq -c$. If $f=-c$, then $e+b$ must be a multiple of $2\pi$, leading to the same conclusion; the two curves are the same and every value of $x$ is a solution.

Lastly, when $a\neq d$ there are a whole world of different possibilities. In this case, a numeric solution may not only be the only approach, but it would be the simplest.

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