Question

Sinx≤√2\2 решение тригонометрических неравенств

Answer 1

Sinx≤√2\2 
3π/4+2πл≤x≤9π/4+2πk,k∈z

Answer 2

Решение тригонометрич#### Solving the Trigonometric Inequality Sin(x) ≤ √2/2

To solve the trigonometric inequality sin(x) ≤ √2/2, we can use the properties of the sine function and the unit circle to find the values of x that satisfy the inequality.

The inequality sin(x) ≤ √2/2 can be rewritten as sin(x) ≤ 1/√2. This inequality represents the values of x for which the sine function is less than or equal to the value of 1/√2.

To find the solutions for this inequality, we can consider the values of x for which the sine function is less than or equal to 1/√2.

The solutions will be in the form of intervals on the real number line.

Solution Steps:

1. Identify the Critical Points: - The critical points for the sine function occur at the values of x where sin(x) = 1/√2. - The values of x for which sin(x) = 1/√2 can be found using the properties of the unit circle and the reference angles for sine.

2. Determine the Intervals: - Once the critical points are identified, we can determine the intervals on the real number line where sin(x) is less than or equal to 1/√2.

3. Express the Solution: - The solution will be expressed as a union of intervals on the real number line that satisfy the inequality sin(x) ≤ 1/√2.

Solution:

The critical points for the sine function occur at the values of x where sin(x) = 1/√2. Using the properties of the unit circle and the reference angles for sine, we find that sin(π/4) = 1/√2.

Therefore, the solution to the inequality sin(x) ≤ 1/√2 is: x ∈ [2nπ + (-1)^n * π/4, 2nπ + (-1)^(n+1) * π/4], where n is an integer.

This solution represents the intervals on the real number line where sin(x) is less than or equal to 1/√2.

I hope this helps! If you have further questions or need additional assistance, feel free to ask.