Find the critical numbers of the function. (Enter your answers as a comma-separated list.)h(x) = sin2(x) + cos(x)    0 < x < 2𝜋

To find the critical numbers of the function h(x) = sin^2(x) + cos(x), we first need to find the derivative of the function.

The derivative of sin^2(x) is 2sin(x)cos(x) using the chain rule, and the derivative of cos(x) is -sin(x).

So, the derivative h'(x) of the function h(x) is:

h'(x) = 2sin(x)cos(x) - sin(x)

We can factor out sin(x) from the equation:

h'(x) = sin(x)(2cos(x) - 1)

The critical numbers are the solutions to the equation h'(x) = 0.

So, we set the equation to 0 and solve for x:

sin(x)(2cos(x) - 1) = 0

This gives us two equations:

sin(x) = 0 and 2cos(x) - 1 = 0

Solving these equations gives us the critical numbers.

For sin(x) = 0, the solutions in the interval 0 < x < 2π are x = π.

For 2cos(x) - 1 = 0, we get cos(x) = 1/2. The solutions in the interval 0 < x < 2π are x = π/3 and x = 5π/3.

So, the critical numbers of the function h(x) = sin^2(x) + cos(x) in the interval 0 < x < 2π are x = π, π/3, 5π/3.