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I have a function in the form of:
$$\cos^{-1} \left(\dfrac{a^2+ bx^2}{2abx}\right) + \cos^{-1} \left(\dfrac {c^2 + dx^2}{2cdx}\right) = e$$

How do I solve for $x$ if all other variables ($a$ through $e$) are known?

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I suppose that it should be $b^2x^2$ and $d^2x^2$ instead of $bx^2$ and $dx^2$ – pedja Jan 15 '12 at 9:35
Have you tried taking the cosine of both sides, then using $\cos{(x+y)}=\cos{x}\cos{y}-\sin{x}\sin{y}$ – Mike Jan 15 '12 at 9:46
@Mike, as far as I know I would end up with something like $$\cos(cos^{-1}(A) + cos^{-1}(B)) = cos(e)$$ and I'm not sure how to solve this. – Adrien Hingert Jan 15 '12 at 9:50
up vote 5 down vote accepted

Building on Mike's suggestion, you can take the cosine of both sides and exploit the fact that $\cos{(x+y)}=\cos{x}\cos{y}-\sin{x}\sin{y}$. Starting with your equation, I make the following substitutions: $$u=\dfrac{a^2+ bx^2}{2abx}, v=\dfrac {c^2 + dx^2}{2cdx}$$ to get the equation $$\cos^{-1}u+\cos^{-1}v=e$$ and take the cosine of both sides: $$\cos(\cos^{-1}u+\cos^{-1}v)=\cos{e}$$ which can be simplified in steps. First, we use Mike's suggestion to get $$\cos(\cos^{-1}u)\cos(\cos^{-1}v)-\sin(\cos^{-1}u)\sin(\cos^{-1}v)=\cos{e}$$ which simplifies to $$uv-\sin(\cos^{-1}u)\sin(\cos^{-1}v)=\cos{e}$$ but from here it is not obvious how to proceed, so we have to get tricky. The key is that $\sin(\cos^{-1}a)=\sqrt{1-a^2}$ for any $a$, so our equation becomes $$uv-\sqrt{1-u^2}\sqrt{1-v^2}=\cos{e}$$ which we rearrange as $$\sqrt{1-u^2}\sqrt{1-v^2}=uv-\cos{e}$$ and square both sides to get $$(1-u^2)(1-v^2)=(uv-\cos{e})^2$$ which is just a polynomial. Substituting the expressions for $u$ and $v$ in terms of $x$ and expanding this out will let you find the solution using the standard methods for solving polynomial equations, although the equation will have degree $8$ so you will need a computer or good calculator to solve it. Be careful to check the solutions you get this way, as sometimes this method may introduce "extraneous" solutions which do not actually solve your equation.

Edit: I forgot to mention that you will get a rational equation rather than a polynomial equation immediately after substituting for $u$ and $v$, so you will have to multiply by the denominators to get a polynomial equation (which will have higher degree, I think 12). Be careful of cases where the denominators are equal to $0$, as these will appear to be solutions but are actually points at which one side of the equation is undefined.

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Building on whose suggestion now? :) – Mike Jan 15 '12 at 10:11
@Mike Oops, misread the comments, off-by-1 error. Editing. – Alex Becker Jan 15 '12 at 10:15
@AlexBecker, thanks this looks great although it will be a real pain to solve :) – Adrien Hingert Jan 15 '12 at 11:07
what are those $x$ and $y$ hanging around when you rearrange the equation? I am sure they are $u$ and $v$. – user21436 Jan 15 '12 at 12:09
@KannappanSampath Yes, they were left over from my original use of $x,y$ before I realized that the OP's equation already used $x$. – Alex Becker Jan 15 '12 at 20:00

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