System of equations with trig functions Can someone explain to me how to solve this system of equations for the parameters $a$, $k$ and $\phi$? I already tried solving the last equation for a and plugging it into another equation, but I guess that doesn't work.
$0=a\sin(5k-\phi)$
$0=a\sin(23k-\phi)$
$0.3=a\sin(-\phi)$
 A: From the third equation, you see that $a$ has to be non-zero. Notice then that the first two equations
$$0 = a\sin(5k-\phi),$$
$$0 = a\sin(23k-\phi),$$
do not specify $a$ at all, because whatever $a$ is, you must have that
$$\sin(5k-\phi) = 0,$$
$$\sin(23k-\phi) = 0.$$
When is $\sin(x)=0$? It happens exactly when $x$ is a multiple of $\pi$, so you get that
$$5k-\phi = n\pi,$$
$$23k-\phi = m\pi,$$
for some integers $m,n\in\mathbb{Z}$, so let's say that some pair $(m,n)$ has been chosen. Subtracting the first equation from the second, we get
$$18k = (m-n)\pi,$$
so
$$k = \frac{m-n}{18}\pi.$$
Plug this into the first equation to get
$$\phi = \frac{5}{18}(m-n)\pi-n\pi.$$
Concerning the third equation
$$0.3 = a\sin(-\phi),$$
you see that we can simply choose $a$ to make it all fit, so
$$a = \frac{0.3}{\sin(-\phi)}.$$
A: Hint:
Realize that the first two statements mean that sin$(5k-\phi)$ and sin$(5k-\phi)$ are both equal to $0$. Thus, $(23k-\phi)$ and $(5k-\phi)$ are both multiples of $\pi$.
$a$ cannot equal 0; if it did the third equation would never be true.
A: Note that solution is not unique, you will likely get a whole family of solutions. 
It follows from the third equation that $a$ is nonzero. Therefore we can cancel it out in the first and second equation, obtaining
$\sin(5k - \phi) = 0$ and $\sin(23k - \phi) = 0$.
Sine has zeros, when its argument is an integer multiple of $\pi$:
$$\begin{align*}
 5k - \phi &= n\pi \\
 23k - \phi &= m\pi \\
\end{align*}$$
where $n$ and $m$ are arbitrary integers.
Subtracting the equations above, we get $18k = (m-n) \pi$, $m-n$ is an arbitrary integer again, call it $r$.
So $k = r \frac{\pi}{18}$. Express $\phi$ from one of the equations above (say the first one) and get
$\phi = 5k - n\pi = 5r \frac{\pi}{18} - n \pi$.
Insert both of these in the last equation and get $$a = \frac{3}{10 \sin(n\pi - \frac{5r\pi}{18})} = \frac{3}{(-1)^{n+1}10\sin(r \frac{\pi}{18})}$$
As far as I know $\sin(\frac{\pi}{18})$ is not a nice value (though, looks like it has an interesting alternate representation see here.
