Question

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

Answer 1

Answer:

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

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

$x=\frac{\pi}{2}$

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:

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