Question

Ctgx cosx+sinx >=2 решения на промежутке [-пи/2;пи/8]

Answer 1

ctgx cosx+sinx >=2 решения на промежутке [-пи/2;пи/8]

Answer 2

To solve the inequality ctgx cosx + sinx >= 2 on the interval [-π/2, π/8], we can follow these steps:

Step 1: Find the critical points. Take the derivative of the left-hand side of the inequality to find critical points where the derivative is equal to zero or undefined.

First, let's simplify the left-hand side of the inequality: ctgx cosx + sinx >= 2 (cosx/sinx) cosx + sinx >= 2 cos²x/sinx + sinx >= 2

Now, find the derivative of the left-hand side: d/dx (cos²x/sinx + sinx) = 0 (-2sinxcosx/sinx²) + cosx = 0 -2cosx/sinx + cosx = 0 cosx(-2/sinx + 1) = 0

To find the critical points, solve -2/sinx + 1 = 0: -2/sinx + 1 = 0 -2 + sinx = 0 sinx = 2

However, sinx cannot be equal to 2 since the range of sinx is [-1, 1]. Therefore, there are no critical points within the given interval.

Step 2: Test the endpoints. Evaluate the left-hand side of the inequality at the endpoints of the interval.

For x = -π/2: ctg(-π/2) cos(-π/2) + sin(-π/2) = 0 - 1 = -1 Since -1 < 2, the inequality is not satisfied at x = -π/2.

For x = π/8: ctg(π/8) cos(π/8) + sin(π/8) ≈ 0.765 > 2 Since 0.765 > 2, the inequality is not satisfied at x = π/8.

Step 3: Analyze the inequality. Since there are no critical points within the interval and the inequality is not satisfied at the endpoints, there are no solutions to the inequality ctgx cosx + sinx >= 2 on the interval [-π/2, π/8].