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 properties

The sine function ( y = \sin x ) has a period of ( 2\pi ), and ( \sin x=- 1 ) when ( x = \frac{3\pi}{2}+2k\pi ), where ( k\in\mathbb{Z} ) (integers).

Step2: Find x in the interval ( (2\pi,4\pi) )

We need to find ( k ) such that ( 2\pi<\frac{3\pi}{2}+2k\pi < 4\pi ). First, subtract ( \frac{3\pi}{2} ) from all parts: ( 2\pi-\frac{3\pi}{2}<2k\pi<4\pi - \frac{3\pi}{2} ) Simplify: ( \frac{\pi}{2}<2k\pi<\frac{5\pi}{2} ) Divide all parts by ( 2\pi ): ( \frac{1}{4}<k < \frac{5}{4} ) Since ( k ) is an integer, ( k = 1 ).

Step3: Calculate x

Substitute ( k = 1 ) into ( x=\frac{3\pi}{2}+2k\pi ): ( x=\frac{3\pi}{2}+2\times1\times\pi=\frac{3\pi + 4\pi}{2}=\frac{7\pi}{2} ) So the numerator is 7 and the denominator is 2.

Answer:

The x - value is (\frac{7\pi}{2}), so the numerator is 7 and the denominator is 2. So the answer is (\frac{7}{2}) (in the form (\frac{[7]\pi}{[2]})).