Question

Сos<-√2\2 решить неравенство ​

Answer 1

Минимальное значение функции y = cos(x) равно -1.

В точке y = -√2/2 значение аргумента x = ±3π/4.

По заданию функция меньше -√2/2, поэтому значение аргумента лежит левее и правее точки, где функция равна -1 на величину (π/4).

Ответ: (3/4)π + 2πk < x < (5/4)π + 2πk, k ∈ Z.

Answer 2

To solve the inequality cos(x) < -√2/2, we need to find the values of x that satisfy this condition.

First, let's identify the angles where cos(x) = -√2/2. We know that cos(45°) = cos(π/4) = -√2/2. Additionally, since cosine is an even function, we have cos(315°) = cos(7π/4) = -√2/2.

Now, we can consider the unit circle to find the angles where cos(x) < -√2/2. On the unit circle, cos(x) represents the x-coordinate of the point corresponding to angle x.

For cos(x) < -√2/2, the x-coordinate should be negative and less than -√2/2. This occurs in the third and fourth quadrants.

In the third quadrant, the angles are between 180° and 270° (π and 3π/2). In the fourth quadrant, the angles are between 270° and 360° (3π/2 and 2π).

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

x ∈ (180°, 270°) U (270°, 360°) [in degrees]

x ∈ (π, 3π/2) U (3π/2, 2π) [in radians]