<s>[INST] Finding $a$ and $b$ such that the graph of $y=-2\sin3x+1$ passes through $(-\frac\pi{12},a)$ and $(b,1)$ Please look at the attached image through the link. I already figured out $b$, which I'm pretty sure is $2\pi/3$. However, I'm confused on figuring out $a$. This is what I tried:
\begin{align*}
y & = -2\sin(3x)+1\\
y & = -2\sin\left(3 \cdot -\frac{\pi}{12}\right)+1\\
y & = -2\sin\left(-\frac{\pi}{4}\right)+1
\end{align*}
I used the unit circle to find $\sin\left(-\frac{\pi}{4}\right)$.
\begin{align*}
y & = -2\left(-\frac{\sqrt{2}}{2}\right)+1\\
y & =  \sqrt{2} + 1\\
y & = 2.41421356
\end{align*}
My answer isn't correct, so could anyone please explain how I should solve for $b$? Thanks!

 [/INST] To find b simply let $1 = -2\sin(3x) +1$ and it is when $\sin3x = 0$ for the some $\frac{\pi}{4} < x < \frac{\pi}{2}$
Since $\arcsin(0) \in (0, \pi, 2\pi...) $ you may obtain $x = 0, \frac{\pi}{3}, \frac{2\pi}{3}...$ from which clearly $\frac{\pi}{3}$ is the desired value
To find a is a similiar method, let $a = 1 - 2\sin(-\frac{\pi}{4})$ such that $ a = 1 + \sqrt{2}$
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