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When I evaluate Solve[a==Sin[b*c], b] to rearrange the following for $ b $:

$$ a = \sin(bc) $$

I get the following result from Mathematica:

$$\begin{align*} \left\{\left\{b\to \text{ConditionalExpression}\left[\frac{-\sin ^{-1}(a)+2 \pi c_1+\pi }{c},c_1\in \mathbb{Z}\right]\right\},\right.\left.\left\{b\to \text{ConditionalExpression}\left[\frac{\sin ^{-1}(a)+2 \pi c_1}{c},c_1\in \mathbb{Z}\right]\right\}\right\} \end{align*}$$

It seems far too complicated. Unless I'm making a huge mistake, surely solving the equation for $ b $ would give:

$$ b = \frac{\sin ^{-1}(a)}{c} $$

Am I doing something wrong?

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closed as too localized by Sjoerd C. de Vries, chris, whuber, m_goldberg, rm -rf Mar 30 '13 at 15:51

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Your answer is one possibility. But how about b = (ArcSin[a] + 2 Pi)/c? If you plug that into your defining equation then you get $a$ also.. so there are many answers, that's what Mathematica is telling you.

For example let

b = (ArcSin[a] + 2 Pi)/c

then calculate

Sin[b c]

and the answer is:

a

Similarly for other multiples of Pi.

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  • $\begingroup$ Ah, I see. It looks like I've got a lot of learning ahead of me if I'm going to be able to make the most of Mathematica. $\endgroup$ – Craig Johnston Mar 29 '13 at 15:54
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You are getting a complete solution - and I think it is beautiful:

Reduce[a == Sin[b*c], b, Reals] // TraditionalForm

enter image description here

The above is a bit closer to what you want - but is "correct".

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