Calculate $x$, if $y = a \cdot \sin{[b(x-c)]}+d$ I am not an expert when it comes to trigonometric functions.

I need to calculate value of $x$ for a program.

If $y = a \cdot \sin{[b(x-c)]}+d$ then what will be the formula to calculate $x$?

Please help..
So this equation $y = a\cdot \sin[b(x-c)]+d$ can be rearranged as

$$\frac{y-d}{a}= \sin[b(x-c)]$$

Using the $\arcsin$ this can be rewritten as

$$\arcsin\left(\frac{y-d}{a}\right) + k\cdot 2\pi= b(x-c)$$

with $k$ an integer. This is so because the $\sin$ function is periodic with period $2\pi$ whereas the $\arcsin$ is actually the inverse of a truncated $\sin$.

Rearranging further:

$$c+\frac{1}{b}\arcsin\left(\frac{y-d}{a}\right) + \frac{2k\cdot \pi}{b}= x$$

Now, there is actually another set of solutions which follows from the fact that $\sin(a) = \sin(\pi - a)$. So that we also have

$$\arcsin\left(\frac{y-d}{a}\right) + k\cdot 2\pi= \pi-b(x-c)$$

or after reworking

$$c-\frac{1}{b}\arcsin\left(\frac{y-d}{a}\right) - \frac{(2k-1)\cdot \pi}{b}= x$$

Again, for $k$ being an integer.

Now, you might want to restrict your solution to angles between $[0,2\pi]$, in which case you don't need the entire sets of solutions but just two of them. If you have further restrictions, there can be a unique solution. 

Here's a plot of what the solutions look like graphically:



The red points correspond to my first formula

$$x_k = c+\frac{1}{b}\arcsin\left(\frac{y-d}{a}\right) + \frac{2k\cdot \pi}{b} \; ,$$

the green points correspond to my second one

$$x'_k=c-\frac{1}{b}\arcsin\left(\frac{y-d}{a}\right) - \frac{(2k-1)\cdot \pi}{b} \; .$$
$\arcsin[b(x-c)] = y - d \implies$

$b(x-c) = \sin(y-d) \implies$

$x=\sin(y-d)/b + c$