Find all the solutions of the equation in the interval [0, 2pi)
sin2x-1=0

show work pls

Step-by-step explanation:

We want to find all solutions to the equation:

In the interval [0, 2π).

First, add one to both sides:

Recall that sin(u) equals one whenever u = π/2. This will occur for every rotation. Hence, we can say that u = π/2 + 2nπ where n is an integer.

And in this case, u = 2x. Thus:

Dividing both sides by two yields:

There are two values of x in the interval [0, 2π) given when n = 0 and n = 1:

Any other solutions will be outside our interval.

Therefore, our solutions are:

no not all of them.

step-by-step explanation:

some of them have a square base.