# Find the biggest negative value of a , for which the maximum of f ( x ) = s i

Find the biggest negative value of a , for which the maximum of
$f\left(x\right)=sin\left(24x+\frac{\pi a}{100}\right)$ is at ${x}_{0}=\pi$
The answer is $a=-150$ , but I don't understand the solving way. I would appreciate if you'd help me please
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Colin Moran
Step 1
The problem involves determining the closest maximum of $\mathrm{sin}\left(24x\right)$ 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 $x=\frac{1}{48}\left(4\pi \cdot n+\pi \right)$ 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 )

Layla Velazquez
Step 1
Recall that the general solution of $\mathrm{sin}\theta =1$ is:

So the general solution for finding all maximum values of f(x) is:

In particular, we know that there is some ${n}_{0}\in \mathbb{Z}$ such that ${x}_{0}=\pi$ so:
$\begin{array}{rl}\pi & =\frac{1}{24}\left(\frac{\pi }{2}-\frac{\pi a}{100}+2\pi {n}_{0}\right)\\ 24\pi & =\frac{\pi }{2}-\frac{\pi a}{100}+2\pi {n}_{0}\\ \frac{\pi a}{100}& =\frac{\pi }{2}-24\pi +2\pi {n}_{0}\\ a& =-2350+200{n}_{0}\end{array}$
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 $a=-2350+200\left(11\right)=-150$ , as desired.