Question

Cosx>-1/2 неравенство

Answer 1

Решить неравенство $\cos x > -\dfrac{1}{2}$.

Решение. Нарисуем единичную окружность и нарисуем прямую $x = -\dfrac{1}{2}$, так как функция $\cos \alpha$ отвечает за ось абсцисс.

$\cos x > -\dfrac{1}{2}$ выполняется на участке, нарисованным синим цветом.

Найдем выколотые точки.

$\cos x = -\dfrac{1}{2};\\x = \pm \arccos \bigg(-\dfrac{1}{2} \bigg) + 2\pi n, \ n \in Z;\\x_{1} = \dfrac{2\pi}{3} + 2\pi n, \ n \in Z;\\x_{2} = -\dfrac{2\pi}{3} + 2\pi n, \ n \in Z.$

Следовательно, $-\dfrac{2\pi}{3} + 2\pi n < x < \ \dfrac{2\pi}{3} + 2\pi n, \ n \in Z.$

Ответ: $x \in \bigg (-\dfrac{2\pi}{3} + 2\pi n; \ \dfrac{2\pi}{3} + 2\pi n \bigg), \ n \in Z.$

Answer 2

To solve the inequality cos(x) > -1/2, we can use the unit circle or trigonometric identities.

1. Using the Unit Circle: First, let's identify the angles where cos(x) is greater than -1/2. On the unit circle, the cosine function represents the x-coordinate of a point on the circle.

Since -1/2 lies between -1 and 0 on the unit circle, we need to find the angles where the x-coordinate is greater than -1/2. These angles are in the second and third quadrants.

In the second quadrant, the cosine function is positive, so we have:

π < x < (3π)/2

In the third quadrant, the cosine function is negative, so we have:

(2π)/3 < x < (3π)/2

Therefore, the solution for cos(x) > -1/2 is:

π < x < (3π)/2 or (2π)/3 < x < (3π)/2

2. Using Trigonometric Identities: We can also use trigonometric identities to solve the inequality. One such identity is:

cos(x) > -1/2 is equivalent to x ∈ (2nπ, 2nπ + (2π)/3) ∪ (2nπ + (4π)/3, 2nπ + 2π)

where n is an integer.

This representation accounts for the periodic nature of the cosine function. The intervals (2nπ, 2nπ + (2π)/3) and (2nπ + (4π)/3, 2nπ + 2π) represent the angles where cos(x) is greater than -1/2.

So, the solution for cos(x) > -1/2 is:

x ∈ (2nπ, 2nπ + (2π)/3) ∪ (2nπ + (4π)/3, 2nπ + 2π), where n is an integer.

Note that these solutions are not unique since the cosine function repeats itself periodically.