To determine which function has a period of 2, we need to analyze the period of each given trigonometric function.
1. For [tex]\( y = -4 \sin 2x \)[/tex]:
- The period of the sine function, [tex]\(\sin(kx)\)[/tex], is given by [tex]\(\frac{2\pi}{k}\)[/tex].
- Here, [tex]\( k = 2 \)[/tex].
- Therefore, the period of [tex]\( -4 \sin 2x \)[/tex] is [tex]\[ \frac{2\pi}{2} = \pi. \][/tex]
- This period is [tex]\(\pi\)[/tex], which is not equal to 2.
2. For [tex]\( y = 2 \sin 3x \)[/tex]:
- The period of the sine function, [tex]\(\sin(kx)\)[/tex], is given by [tex]\(\frac{2\pi}{k}\)[/tex].
- Here, [tex]\( k = 3 \)[/tex].
- Therefore, the period of [tex]\( 2 \sin 3x \)[/tex] is [tex]\[ \frac{2\pi}{3} \][/tex].
- This period is [tex]\(\frac{2\pi}{3}\)[/tex], which is not equal to 2.
3. For [tex]\( y = 3 \cos \pi x \)[/tex]:
- The period of the cosine function, [tex]\(\cos(kx)\)[/tex], is given by [tex]\(\frac{2\pi}{k}\)[/tex].
- Here, [tex]\( k = \pi \)[/tex].
- Therefore, the period of [tex]\( 3 \cos \pi x \)[/tex] is [tex]\[ \frac{2\pi}{\pi} = 2 \][/tex].
- This period is 2.
4. For [tex]\( y = \cos \left(x - \frac{\pi}{2}\right) \)[/tex]:
- The period of the cosine function, [tex]\(\cos(x - c)\)[/tex], does not change because horizontal shifting does not affect the period.
- The period of [tex]\( \cos(x) \)[/tex] is [tex]\(2\pi\)[/tex].
- Therefore, the period of [tex]\( \cos \left(x - \frac{\pi}{2}\right) \)[/tex] remains [tex]\( 2\pi \)[/tex].
- This period is [tex]\(2\pi\)[/tex], which is not equal to 2.
So, the function out of the given options which has a period of 2 is:
[tex]\[ \boxed{y = 3 \cos \pi x} \][/tex]
which corresponds to option C.
