Answer :

Sure, let's determine the period of the function [tex]\( y = -2 \sin 3x \)[/tex].

1. Identify the Standard Form:
The general form of a sine function is [tex]\( y = A \sin(Bx + C) + D \)[/tex], where:
- [tex]\( A \)[/tex] is the amplitude.
- [tex]\( B \)[/tex] affects the period of the function.
- [tex]\( C \)[/tex] is the phase shift.
- [tex]\( D \)[/tex] is the vertical shift.

2. Extract the Relevant Parameter:
For our function [tex]\( y = -2 \sin 3x \)[/tex], the amplitude [tex]\( A = -2 \)[/tex], but this does not affect the period. The key parameter here is [tex]\( B \)[/tex], which is 3 in this case.

3. Determine the Period:
The period [tex]\( T \)[/tex] of the sine function is determined by the parameter [tex]\( B \)[/tex]. The formula to find the period is:
[tex]\[ T = \frac{2\pi}{B} \][/tex]
Substituting the value of [tex]\( B = 3 \)[/tex] into the formula gives us:
[tex]\[ T = \frac{2\pi}{3} \][/tex]

4. Simplify the Period:
This result represents the length of one complete cycle of the sine wave.

Thus, the period of the function [tex]\( y = -2 \sin 3x \)[/tex] is [tex]\( \frac{2\pi}{3} \)[/tex].

Through calculation and confirmation, the numerical value of this period is approximately:
[tex]\[ 2.0943951023931953 \][/tex]

So, the period of the function [tex]\( y = -2 \sin 3x \)[/tex] is [tex]\(2.0943951023931953\)[/tex].

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