Question

2cos²x-1=0 2cos²x-cos²x-sin²x=0 cos²x-sin²x=0 cos2x=0 2x=π/2+πk, k∈Z(Как это нашли? Это относится к особым случаям?) x=π/4+πk/2, k∈Z Объясните

Answer 1

Ну 1 разложили на множители, потом привели подобные слагаемые. То, что получилось соьраои в формулу. Нашли где х равен 0,потом ответ разделили на 2, т. к. в уравнении 2х

Answer 2

To solve the equation 2cos²x - 1 = 0, we can follow these steps:

Step 1: Combine like terms on the left side of the equation. 2cos²x - cos²x - sin²x = 0

Step 2: Simplify the equation. cos²x - sin²x = 0

Step 3: Use the trigonometric identity cos²θ - sin²θ = cos(2θ) to rewrite the equation. cos(2x) = 0

Step 4: Find the solutions for cos(2x) = 0.

In the unit circle, the values of θ where cos(θ) = 0 are π/2 + kπ, where k is an integer. So, for our equation cos(2x) = 0, we can set 2x equal to these values:

2x = π/2 + kπ, where k ∈ Z

Step 5: Solve for x. Divide both sides by 2: x = π/4 + kπ/2, where k ∈ Z

Now, let's explain the steps:

1. In Step 1, we simplified the equation by combining like terms, which brought us to cos²x - sin²x = 0.

2. In Step 2, we applied the trigonometric identity cos²θ - sin²θ = cos(2θ), where θ is replaced by x in our case, leading us to cos(2x) = 0.

3. In Step 3, we found the values of x for which cos(2x) equals 0. These values occur when 2x is equal to π/2 + kπ, where k is an integer. This is a result of the periodicity of the cosine function, which repeats every 2π radians.

4. In Step 4, we solved for x by dividing both sides of the equation by 2, which gave us the solutions x = π/4 + kπ/2, where k is an integer.

These solutions represent the values of x that satisfy the original equation 2cos²x - 1 = 0. In this case, they can be considered "special cases" or "particular solutions" because they are based on specific properties of the cosine function and the trigonometric identity, which led us to find a concise and general form for the solutions in terms of k (integer).