The function is given to be:
[tex]y=5\sin3x[/tex]
The zeros of the function refer to the x-values of the function when y = 0. This is shown below:
[tex]5\sin3x=0[/tex]
Divide both sides by 5:
[tex]\sin3x=0[/tex]
Find the sine inverse of both sides:
[tex]3x=\sin^{-1}(0)[/tex]
Recall the sine inverse of 0 is given to be:
[tex]\begin{gathered} \sin^{-1}0=0+2\pi n \\ or \\ \sin^{-1}0=\pi+2\pi n \end{gathered}[/tex]
where n refers to the number of cycles of the function.
Therefore, it can be had that:
[tex]\begin{gathered} 3x=0+2\pi n \\ \therefore \\ x=\frac{2\pi n}{3} \\ OR \\ 3x=\pi+2\pi n \\ \therefore \\ x=\frac{\pi}{3}+\frac{2\pi n}{3} \end{gathered}[/tex]
Substituting 0 and integer values of n into the solutions above, the value of x that will fall within the interval will be:
[tex]\begin{gathered} x=0,\frac{2\pi}{3},\frac{4\pi}{3},2\pi \\ OR \\ x=\frac{\pi}{3},\pi,\frac{5\pi}{3} \end{gathered}[/tex]
Hence, the zeros of the function between the interval [0, 2π] are:
[tex]x=0,\:x=\frac{\pi }{3},\:x=\frac{2\pi }{3},\:x=\pi ,\:x=\frac{4\pi }{3},\:x=\frac{5\pi }{3},\:x=2\pi [/tex]
