Sine Parametric function exercise Find the biggest negative value of $a$ , for which the maximum of 
$f(x) =sin(24x+\frac{πa}{100})$ is at $x_0=π$
The answer is $a=-150$, but I don't understand the solving way. I would appreciate if you'd help me please
 A: The problem involves determining the closest maximum of $\sin(24x)$ to $x_0 = \pi$ and then shift the curve to the right amount to put it in $x_0 = \pi$.
Maximums are at the points $x=\frac{1}{48}(4\pi\cdot n + \pi)$ so you have that the closest maximum is at $n= 11$ so $x_{closest} = \frac{45}{48}\pi$ 
This means that you have to shift the curve to the right by $\frac{3}{48}\pi$ to respect the condition that the maximum is in $x_0 = \pi$. Can you finish from here? (you can use $x = y - \frac{3}{48}\pi$ to shift the curve )
A: Recall that the general solution of $\sin\theta = 1$ is:
$$
\theta = \frac{\pi}{2} + 2\pi n, \text{where }n \in \mathbb Z
$$
So the general solution for finding all maximum values of $f(x)$ is:
\begin{align*}
24x + \frac{\pi a}{100} &= \frac{\pi}{2} + 2\pi n, \text{where }n \in \mathbb Z \\
24x  &= \frac{\pi}{2} - \frac{\pi a}{100} + 2\pi n, \text{where }n \in \mathbb Z \\
x  &= \frac{1}{24} \left(\frac{\pi}{2} - \frac{\pi a}{100} + 2\pi n \right), \text{where }n \in \mathbb Z \\
\end{align*}
In particular, we know that there is some $n_0 \in \mathbb Z$ such that $x_0 = \pi$, so:
\begin{align*}
\pi  &= \frac{1}{24} \left(\frac{\pi}{2} - \frac{\pi a}{100} + 2\pi n_0 \right) \\
24\pi  &= \frac{\pi}{2} - \frac{\pi a}{100} + 2\pi n_0 \\
\frac{\pi a}{100}  &= \frac{\pi}{2} - 24\pi + 2\pi n_0 \\
a  &= -2350 + 200 n_0 \\
\end{align*}
Since $a < 0$, we know that $n_0 < \frac{2350}{200} = 11.75$. Rounding down to the nearest integer, we find that $n_0 = 11$ so that $a = -2350 + 200(11) = -150$, as desired. $~~\blacksquare$
