Rearranging an equation with the sine function in it [closed]

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]


a


Similarly for other multiples of Pi.

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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. –  Craig Johnston Mar 29 '13 at 15:54

You are getting a complete solution - and I think it is beautiful:

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


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

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