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fiasKO [112]
2 years ago
9

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

Mathematics
1 answer:
Assoli18 [71]2 years ago
5 0

Answer:

x=\{0,\frac{\pi}{2},\pi,\frac{3\pi}{2}\}

Step-by-step explanation:

<u>Trigonometric Equations</u>

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=\frac{\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=\frac{3\pi}{2}

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

\mathbf{x=\{0,\frac{\pi}{2},\pi,\frac{3\pi}{2}\}}

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