Question

Решить неравенство sin x/2 < 1/2

Answer 1

sin x/2 < 1/2

-π-arcsin0.5<х/2<arcsin0.5+2πn, где  n∈Z;

-π-π/6<х/2<π/6+2πn, где  n∈Z;

-2π-π/3<х<π/3+4πn, где  n∈Z;

-7π/3<х<π/3+4πn, где  n∈Z;

Answer 2

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

Step 1: Find the critical points where sin(x/2) = 1/2. Step 2: Determine the intervals where sin(x/2) is less than 1/2. Step 3: Express the solution in interval notation.

Step 1: Critical points where sin(x/2) = 1/2 To find the critical points, we need to solve the equation sin(x/2) = 1/2:

sin(x/2) = 1/2

The solutions to this equation can be found by looking at the unit circle or using trigonometric identities. In this case, x/2 must be one of the following angles for which the sine function takes the value of 1/2:

x/2 = π/6 + 2πk or x/2 = 5π/6 + 2πk,

where k is an integer.

Step 2: Determine the intervals where sin(x/2) < 1/2. We need to identify the regions where sin(x/2) is less than 1/2. To do this, we can draw a sine curve on the unit circle or use the information from Step 1.

Since sin(x/2) is a periodic function with a period of 2π, we can analyze the critical points on one full period and then extend the solution to the entire real line.

The critical points we found were: x/2 = π/6 + 2πk or x/2 = 5π/6 + 2πk.

Let's analyze the intervals for k = 0:

For x/2 = π/6, we have x = π/3. For x/2 = 5π/6, we have x = 5π/3.

Thus, the interval for k = 0 is π/3 < x < 5π/3.

Step 3: Express the solution in interval notation. The solution to the inequality sin(x/2) < 1/2 is: x ∈ (π/3, 5π/3).

Keep in mind that since sin(x) is a periodic function, there are infinitely many solutions to this inequality, spaced at intervals of 2π. The above interval (π/3, 5π/3) is just one such interval.