Question

determine the period of the function ( f(x)=2 sin left(3 x+\frac{pi}{3}
ight)+5 ). (1 point) ( \frac{pi}{9} ) ( \frac{2 pi}{3} ) ( 2 pi ) ( pi )

Explanation:

Step1: Recall the formula for the period of a sine function

The general form of a sine function is \(y = A\sin(Bx - C)+D\), and its period \(T=\frac{2\pi}{|B|}\).

Step2: Identify the value of \(B\) in the given function

For the function \(f(x)=2\sin(3x + \frac{\pi}{3})+5\), comparing with \(y = A\sin(Bx - C)+D\), we have \(B = 3\).

Step3: Calculate the period

Substitute \(B = 3\) into the period formula \(T=\frac{2\pi}{|B|}\). So \(T=\frac{2\pi}{3}\).

Answer:

\(\frac{2\pi}{3}\) (corresponding to the second option \( \frac{2\pi}{3}\))