Question

Решите плз срочно 1) sin3x-sin7x=0 2) cos2x-cos6x=0 3) cosx+cos2x=0

Answer 1

Решение во вложении.

Answer 2

1. sin3x - sin7x = 0

Using the identity sin(A) - sin(B) = 2 cos((A + B)/2) sin((A - B)/2), we can write:

sin3x - sin7x = 2 cos(5x/2) sin(-2x/2) = -2 cos(5x/2) sin(x)

Therefore, the equation sin3x - sin7x = 0 can be written as:

-2 cos(5x/2) sin(x) = 0

This equation has two solutions:

cos(5x/2) = 0 or sin(x) = 0

For cos(5x/2) = 0, we have:

5x/2 = (2n + 1)π/2, where n is an integer

x = (2n + 1)π/5

For sin(x) = 0, we have:

x = nπ, where n is an integer

Therefore, the solutions to the equation sin3x - sin7x = 0 are:

x = nπ or x = (2n + 1)π/5, where n is an integer.

2. cos2x - cos6x = 0

Using the identity cos(A) - cos(B) = -2 sin((A + B)/2) sin((A - B)/2), we can write:

cos2x - cos6x = -2 sin(4x/2) sin(-2x/2) = 2 sin(2x) sin(x)

Therefore, the equation cos2x - cos6x = 0 can be written as:

2 sin(2x) sin(x) = 0

This equation has two solutions:

sin(2x) = 0 or sin(x) = 0

For sin(2x) = 0, we have:

2x = nπ, where n is an integer

x = nπ/2

For sin(x) = 0, we have:

x = nπ, where n is an integer

Therefore, the solutions to the equation cos2x - cos6x = 0 are:

x = nπ/2 or x = nπ, where n is an integer.

3. cosx + cos2x = 0

Using the identity cos(A) + cos(B) = 2 cos((A + B)/2) cos((A - B)/2), we can write:

cosx + cos2x = 2 cos(5x/4) cos(-x/4)

Therefore, the equation cosx + cos2x = 0 can be written as:

2 cos(5x/4) cos(-x/4) = 0

This equation has two solutions:

cos(5x/4) = 0 or cos(-x/4) = 0

For cos(5x/4) = 0, we have:

5x/4 = (2n + 1)π/2, where n is an integer

x = (2n + 1)π/10

For cos(-x/4) = 0, we have:

-x/4 = (2n + 1)π/2, where n is an integer

x = -(2n + 1)2π

Therefore, the solutions to the equation cosx + cos2x = 0 are:

x = (2n + 1)π/10 or x = -(2n + 1)2π, where n is an integer.