Question

4cos²x - 4cos x + 1 = 0

Answer 1

Ответ:(0,4-4cos(1))

Объяснение:

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Answer 2

Ответ:

x=±π/3+2kπ, k∈Z

Объяснение:

4cos²x - 4cos x + 1 = 0

(2cosx)² - 2·1·(2cos x) + 1² = 0

(2cosx-1)²=0

2cosx-1=0

2cosx=1

cosx=0,5

x=±arccos0,5+2kπ

x=±π/3+2kπ, k∈Z

Answer 3

To solve the equation 4cos²x - 4cosx + 1 = 0, we can use some algebraic techniques. First, notice that the equation is in the form of a quadratic equation. We can let cosx be our variable, and then solve for it using the quadratic formula.

The quadratic formula is given by:

x = (-b ± √(b² - 4ac)) / 2a

In our equation, a = 4, b = -4, and c = 1. Plugging these values into the quadratic formula:

cosx = [4 ± √((-4)² - 4 * 4 * 1)] / (2 * 4)

cosx = [4 ± √(16 - 16)] / 8

cosx = [4 ± √0] / 8

Now, since the discriminant (b² - 4ac) is zero, there is only one real root for cosx:

cosx = 4/8 = 1/2

Now, we need to find the angles where cosx is equal to 1/2. For that, we look at the unit circle or use a calculator:

cosx = 1/2 is true for x = π/3 or x = 5π/3 (and other equivalent angles).

So, the solutions for the original equation 4cos²x - 4cosx + 1 = 0 are:

x = π/3 + 2πn, and x = 5π/3 + 2πn, where n is an integer.