Question

√2sin(pi/2-x)< 1.......

Answer 1

2 косин +3*:#5:*73$_68¥998 (;:=%

Answer 2

To solve the inequality √2sin(pi/2-x) < 1, we can follow these steps:

1. Simplify the expression on the left side: √2sin(pi/2-x).

   • The square root of 2 is a positive constant, so it does not affect the inequality.
   • We can rewrite sin(pi/2 - x) as cos(x) using the complementary angle property.

   The inequality becomes: cos(x) < 1.

2. Since cosine values range from -1 to 1, we know that cos(x) ≤ 1 for all values of x. So, cos(x) < 1 is true for all x, except when cos(x) = 1.

3. To find the values of x where cos(x) = 1, we need to consider the unit circle.

   • Cosine is equal to 1 at x = 0, 2π, 4π, 6π, and so on (where x is an integer multiple of 2π).

Therefore, the solution to the inequality √2sin(pi/2-x) < 1 is:

x ∈ (-∞, 0) ∪ (0 + 2kπ, 2π + 2kπ), where k is an integer.