In general a trigonometric function of the form:
![y=A\sin (B(x+C))+D](https://img.qammunity.org/2023/formulas/mathematics/college/ii36i37a7vtsrezkio0p5hxghf9h69s2y7.png)
has the following properties:
• A is the amplitude.
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• C is the phase shift (positive to the left)
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• D is the vertical shift
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• The period is given as:
![(2\pi)/(B)](https://img.qammunity.org/2023/formulas/physics/college/6kecy9rvg280c84x7xmng8be0hti1ywz42.png)
From the information given we have that the amplitude is 3, then:
![A=3](https://img.qammunity.org/2023/formulas/mathematics/college/ng3u4mov8h8s4lnasow56uiexfvna1p0r8.png)
The period is 6pi then we have:
![\begin{gathered} 6\pi=(2\pi)/(B) \\ B=(2\pi)/(6\pi) \\ B=(1)/(3) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/b1yu78lzg32gwxi4lpqz6fljjtpiyyr50n.png)
The horizontal shift is:
![C=-(3\pi)/(2)](https://img.qammunity.org/2023/formulas/mathematics/college/fwg8e06r7swz9gm3ss6kmgwxejifzu87uy.png)
And the vertical shift is:
![D=-1](https://img.qammunity.org/2023/formulas/mathematics/college/purjbgwkeqh9cdcqs9b6rzw8pvtnyxsfw7.png)
Once we know the values we plug them in the general expression for the sine function, our function is:
![y=3\sin ((1)/(3)(x-(3\pi)/(2)))-1](https://img.qammunity.org/2023/formulas/mathematics/college/1zv7g9pvw770n02r1rh4qrm2mv4fk0j815.png)
Now that we have the function we can find the its value when x=2pi, plugging this value of x in the expression we have:
![\begin{gathered} y=3\sin ((1)/(3)(2\pi-(3\pi)/(2)))-1 \\ y=3\sin ((1)/(3)((\pi)/(2)))-1 \\ y=3\sin ((\pi)/(6))=-1 \\ y=3((1)/(2))-1 \\ y=(3)/(2)-1 \\ y=(1)/(2) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/3n9a7l4qfjkahbimq3riezrm508wpuznj0.png)
Therefore, the value of y for the given x is 1/2