Question

При каких значениях x верно равенство y' * y + y^2 =0,если y = 2sinx?​

Answer 1

Ответ:

Объяснение:

вот и все

Answer 2

To find the values of x for which the equation y' * y + y^2 = 0 is satisfied, where y = 2sinx, we first need to find the derivative of y with respect to x (y').

Given y = 2sinx, the derivative y' can be found using the chain rule:

y' = d(2sinx)/dx = 2cosx.

Now, let's substitute y and y' back into the equation:

y' * y + y^2 = 2cosx * 2sinx + (2sinx)^2 = 4cosx * sinx + 4sin^2x = 4sinx(cosx + sinx).

Now we need to solve the equation 4sinx(cosx + sinx) = 0 for x:

1. sinx = 0: If sinx = 0, then x can take the values of 0, π, 2π, 3π, etc.

2. cosx + sinx = 0: We can write cosx = -sinx: cos^2x + sin^2x = (-sinx)^2 1 + sin^2x = sin^2x 1 = 0 (This is not possible).

Therefore, there are no additional solutions for x when cosx + sinx = 0.

So, the values of x for which the equation y' * y + y^2 = 0 is satisfied when y = 2sinx are: x = 0, π, 2π, 3π, etc.