suppose y = sin x. what is the x - value between 2π and 4π corresponding to a y - value of - 1?

Explanation:

Step1: Recall sine - function property

The general solution of the equation $\sin x=- 1$ is $x=\frac{3\pi}{2}+2k\pi$, where $k\in\mathbb{Z}$.

Step2: Find the value of $k$ for the given interval

We want to find $x$ in the interval $(2\pi,4\pi)$. Let $\frac{3\pi}{2}+2k\pi\in(2\pi,4\pi)$. First, solve the left - hand side of the inequality $2\pi<\frac{3\pi}{2}+2k\pi$. Subtract $\frac{3\pi}{2}$ from both sides: $2\pi-\frac{3\pi}{2}<2k\pi$, $\frac{4\pi - 3\pi}{2}<2k\pi$, $\frac{\pi}{2}<2k\pi$, $\frac{1}{4}<k$. Then, solve the right - hand side of the inequality $\frac{3\pi}{2}+2k\pi<4\pi$. Subtract $\frac{3\pi}{2}$ from both sides: $2k\pi<4\pi-\frac{3\pi}{2}$, $2k\pi<\frac{8\pi - 3\pi}{2}$, $2k\pi<\frac{5\pi}{2}$, $k < \frac{5}{4}$. Since $k\in\mathbb{Z}$, then $k = 1$.

Step3: Calculate the $x$ value

When $k = 1$, $x=\frac{3\pi}{2}+2\pi=\frac{3\pi + 4\pi}{2}=\frac{7\pi}{2}$.

Answer:

$\frac{7}{2}$