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

Step-by-step explanation:

Trigonometric Equations

Solve

for x in [0,2π].

Dividing by 3:

Solving for 2x by using the inverse sine function:

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

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

Again, dividing by 2:

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

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step-by-step explanation: there is no picture

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