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Solve for $x$ in terms of the constants $n$ and $A$.

$$\sqrt{1-(\frac{\sin x}{n})^2}\sqrt{1-(n\sin (A- \arcsin (\frac{\sin x}{n})))^2}=\cos x \cos (A- \arcsin (\frac{\sin x}{n}))$$

This gets really messy when I try and simplify and manipulate.

I notice that there is a pair of difference-of-two-squares on the left. But multiplying out just makes things even messier.

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Define $s := \sin x$ and $r := n^2-s^2$.

Then, $$\sin\left(A -\operatorname{asin}\frac{s}{n}\right) = \sin A \cos\operatorname{asin}\frac{s}{n}-\cos A \sin \operatorname{asin}\frac{s}{n} = \frac{1}{n}\left(r \sin A - s \cos A\right)$$ $$\cos\left(A -\operatorname{asin}\frac{s}{n}\right) = \cdots = \frac{1}{n}\left(r \cos A + s \sin A\right)$$

Squaring your equation, and multiplying-through by $n^2$, gives

$$\left(n^2-s^2\right)\left(1 - \left( r \sin A - s \cos A \right)^2\right) = \left( 1 - s^2 \right)( r \cos A + s \sin A )^2$$

With a bit of symbol-crunching help from Mathematica, this simplifies to $$(n^2-1) \sin A \cdot \left(-n^2 \sin A + 2 s^2 \sin A + 2 r s \cos A\;\right) = 0$$

For $\sin A \neq 0$ and $n \neq \pm 1$, we can eliminate $r$ to get

$$\left(\;n^2 ( 1 - \cos A ) - 2 s^2\;\right)\; \left(\;n^2 ( 1 + \cos A ) - 2 s^2\;\right) = 0$$ which leads to $$\begin{align} s^2 &= n^2 \frac{1-\cos A}{2} = n^2 \sin^2 \frac{A}{2} \quad\to\quad \sin x = \pm n \sin\frac{A}{2}\\[6pt] s^2 &= n^2 \frac{1+\cos A}{2} = n^2 \cos^2 \frac{A}{2} \quad\to\quad \sin x = \pm n \cos\frac{A}{2} \end{align}$$

Sanity checking for extraneous solutions to the original equation is left as an exercise to the reader.

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