How do you solve the equation 0 = 2 sin(2x) cos(2x) on the interval [0, π]?

To solve the equation 0 = 2 sin(2x) cos(2x), we can start by simplifying it using a trigonometric identity.

Recall that sin(2x) = 2 sin(x) cos(x). This allows us to rewrite the equation:

0 = sin(4x)

Now we can see that the equation is set to zero, which means we are looking for the angles 4x for which sin is zero.

This occurs at every integer multiple of π:

4x = nπ, where n is any integer.

Now we will solve for x:

x = nπ/4

We’re interested in the values of x that fall within the interval [0, π]. Going through the values of n:

• If n = 0, then x = 0
• If n = 1, then x = π/4
• If n = 2, then x = π/2
• If n = 3, then x = 3π/4
• If n = 4, then x = π

Thus, the solutions to the equation 0 = 2 sin(2x) cos(2x) in the interval [0, π] are:

• x = 0
• x = π/4
• x = π/2
• x = 3π/4
• x = π