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!


You have the correct value for $a$. Perhaps it was marked incorrect if you entered your approximation rather than the exact value of $\sqrt{2}+1$.
For $b$, you can simply let $y=1$ and solve:
$$y=-2\sin(3x)+1$$
$$1=-2\sin(3b)+1$$
$$-2\sin(3b)=0$$
$$\sin(3b)=0$$
In your diagram, it looks like $\frac{\pi}{4}<b<\frac{\pi}{2}$. Use your unit circle to determine which values of $3b$ produce $\sin(3b)=0$, and find a value of $b$ that satisfies both $\sin (3b) = 0$ and $\frac{\pi}{4}<b<\frac{\pi}{2}$. Your answer of $b=\frac{2\pi}{3}$ satisfies the first criterion, but not the second.
