Question

2cos(4x+pi/2)=корень из 3 Sin x^2=0 Помогите пожалуйста

Answer 1

Ответ:

решение на фотографии

Answer 2

Solving the Equation 2cos(4x+π/2) = √3 and Sin(x^2) = 0

To solve the equation 2cos(4x+π/2) = √3, we can use the trigonometric identity cos(2θ) = 1 - 2sin^2(θ) to simplify the equation.

For 2cos(4x+π/2) = √3: Using the identity cos(2θ) = 1 - 2sin^2(θ), we get: 2cos(4x+π/2) = 2cos(2(2x+π/4)) = 2(1 - 2sin^2(2x+π/4)) = √3

Solving for sin(2x+π/4), we have: 1 - 2sin^2(2x+π/4) = √3/2 2sin^2(2x+π/4) = 1 - √3/2 sin^2(2x+π/4) = (1 - √3/2)/2 sin(2x+π/4) = ±√((1 - √3)/4)

For Sin(x^2) = 0: The equation Sin(x^2) = 0 implies that x^2 = nπ, where n is an integer.

Combining the two equations, we have: x^2 = nπ 2x+π/4 = arcsin(±√((1 - √3)/4))

Solving for x, we get: x = ±√(nπ - π/4) + 2nπ

These are the solutions for the given equations.

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