Question

4cos^2 2x=0 хелп плиз

Answer 1

Ответ:

Смотри разбор

Пошаговое объяснение:

Сразу можно поделить на 4

cos(2x)^2=0

Результатом возведения в степень может быть только 0 тогда, когда основание равно 0

cos(2x)=0

2x= pi/2 + pi*n n∈z

x = pi/4 + pi*n/2  n∈z

Answer 2

To solve the equation 4cos^2(2x) = 0, follow these steps:

Step 1: Recall the identity cos^2(θ) = 1 - sin^2(θ).

Step 2: Substitute this identity into the equation: 4(1 - sin^2(2x)) = 0

Step 3: Distribute the 4: 4 - 4sin^2(2x) = 0

Step 4: Divide both sides by 4: 1 - sin^2(2x) = 0

Step 5: Rearrange the equation: sin^2(2x) = 1

Step 6: Take the square root of both sides: sin(2x) = ±1

Step 7: Solve for 2x: 2x = arcsin(±1)

Remember that the sine function's range is [-1, 1], so only the values ±1 are possible. This leads to two cases:

For sin(2x) = 1: 2x = π/2 + 2πn, where n is an integer.

For sin(2x) = -1: 2x = -π/2 + 2πn, where n is an integer.

Step 8: Solve for x: x = (π/4) + πn, for the case of sin(2x) = 1. x = (-π/4) + πn, for the case of sin(2x) = -1.

These equations provide the general solutions for the equation 4cos^2(2x) = 0, where n is an integer. Each solution corresponds to a value of x that satisfies the original equation.