Question

Where are the zeros for f(x)=3sin2x on the interval [0,2π]?

Answer 1

The zeros for f(x) = 3sin(2x) on the interval [0, 2π] are:

x = 0

x = π

x = 2π

Step-by-step explanation:

We know that the sine function has zeros at multiples of π. Since 2x is doubled inside the sine function, the zeros of f(x) occur at half the intervals of the regular sine function.

Zero at x = 0:

sin(2 * 0) = sin(0) = 0

Therefore, f(0) = 3sin(2 * 0) = 0.

Zero at x = π:

sin(2 * π) = sin(2π) = 0

Therefore, f(π) = 3sin(2 * π) = 0.

Zero at x = 2π:

sin(2 * 2π) = sin(4π) = 0

Therefore, f(2π) = 3sin(2 * 2π) = 0.

There are no other zeros for f(x) on the interval [0, 2π] because in this interval, sin(2x) never takes on values outside the range [-1, 1], and therefore, 3sin(2x) never reaches 0 except at the points mentioned above.

Therefore, the zeros of f(x) on the interval [0, 2π] are x = 0, x = π, and x = 2π.

Answer 2

`x=\{0,(\pi)/(2),\pi,(3\pi)/(2)\}`

Step-by-step explanation:

Trigonometric Equations

Solve

`3\sin 2x=0`

for x in [0,2π].

Dividing by 3:

`\sin 2x=0`

Solving for 2x by using the inverse sine function:

`2x=\arcsin (0)`

There are two angles in the first turn of the trigonometric circle whose sine is 0:

2x=0

2x=π

Dividing by 2, we get the first two solutions:

x = 0

`x=(\pi)/(2)`

Since the argument of the sine is double, we look for solutions in the next turn of the circle, that is:

`2x=2\pi`

`2x=3\pi`

Again, dividing by 2:

`x=\pi`

`x=(3\pi)/(2)`

No more solutions can be found, thus the solutions are:

`\mathbf{x=\{0,(\pi)/(2),\pi,(3\pi)/(2)\}}`