Question

F'(x) = sin2x-1 f'(x)=0 sin2x-1=0sin2x=1 дальше ?

Answer 1

2х=пи/2+2пиn
X=пи/4+пиn

Answer 2

Looks like you're working with derivatives and trigonometric functions. Let's break down your steps and see where we can go from there:

1. You have the derivative of a function F(x), which is F'(x) = sin(2x) - 1.
2. You want to find the points where the derivative is equal to 0, so you set f'(x) = 0: sin(2x) - 1 = 0.
3. Solving for sin(2x) = 1: This would mean that 2x = π/2 + 2πn, where n is an integer, because the sine function has a period of 2π and its maximum value is 1.
4. Solving for x: Divide both sides by 2: x = π/4 + πn, where n is an integer.

So, you have found that the values of x where the derivative F'(x) is equal to 0 are given by x = π/4 + πn, where n can be any integer.

If you are looking for critical points of the original function F(x), you need to find the x-values where the derivative F'(x) is either 0 or undefined. Since you've found that F'(x) is 0 at x = π/4 + πn, these are the critical points of the function F(x).