Question

А) sinx >1/2 Cosx<0 Sinx>0 Cosx<-1/2

Answer 1

Ответ:

Объяснение:1)sinx>1/2 ⇔π/6+2πn<x<5π/6+2πn, n∈z

ответ:(π/6+2πn;5π/6+2πn),n∈z

2)cosx<0 ⇔π/2+2πn<x<3π/2+2πn,n∈z    ответ:(π/2+2πn;3π/2+2πn),n∈z

3)sinx>0⇔2πn<x<π+2πn,n∈z  ответ; (2πn;π+2πn),n∈z

4)cosx<-1/2,  2π/3+2πn<x<4π/3+2πn,n∈z

Answer 2

To solve the system of inequalities:

A) sin(x) > 1/2 cos(x) < 0

First, let's analyze sin(x) > 1/2. This means that the value of sin(x) is greater than 1/2.

In the unit circle, sin(x) represents the y-coordinate of a point on the circle at angle x. For sin(x) > 1/2, we need to find the angles x where the y-coordinate is greater than 1/2.

The angles in the first and second quadrants have positive y-coordinates. In the first quadrant (0° to 90°), sin(x) is increasing from 0 to 1. In the second quadrant (90° to 180°), sin(x) is decreasing from 1 to 0.

Thus, the solution for sin(x) > 1/2 is x ∈ (0°, 90°) ∪ (270°, 360°).

Next, let's consider cos(x) < 0. This means that the value of cos(x) is negative.

In the unit circle, cos(x) represents the x-coordinate of a point on the circle at angle x. For cos(x) < 0, we need to find the angles x where the x-coordinate is negative.

The angles in the second and third quadrants have negative x-coordinates. In the second quadrant (90° to 180°), cos(x) is negative. In the third quadrant (180° to 270°), cos(x) is also negative.

Thus, the solution for cos(x) < 0 is x ∈ (90°, 180°) ∪ (180°, 270°).

To find the common solution to both inequalities, we need to find the intersection of the solution sets.

The intersection of x ∈ (0°, 90°) ∪ (270°, 360°) and x ∈ (90°, 180°) ∪ (180°, 270°) is x ∈ (270°, 360°).

Therefore, the solution to the system of inequalities sin(x) > 1/2 and cos(x) < 0 is x ∈ (270°, 360°).