Cos^2 3x-cos3x*cos5x=0

Предыдущий
вопрос Следующий
вопрос

To solve cos^2(3x) - cos(3x)cos(5x) = 0, we can first factor out a common factor of cos(3x) from both terms:

cos(3x)[cos(3x) - cos(5x)] = 0

Now we have a product of two factors that equals zero. This means that at least one of the factors must be zero. So we have two possibilities:

1) cos(3x) = 0
2) cos(3x) - cos(5x) = 0

1) To solve cos(3x) = 0, we can find the values of x that satisfy this equation. Since cos(3x) = 0 when 3x = (2n + 1)π/2 where n is an integer, then:

3x = (2n + 1)π/2
x = (2n + 1)π/6

2) To solve cos(3x) - cos(5x) = 0, we can use the trigonometric identity:

cos(a) - cos(b) = -2sin[(a + b)/2]sin[(a - b)/2]

Applying this formula to the equation cos(3x) - cos(5x) = 0, we get:

-2sin[(3x + 5x)/2]sin[(3x - 5x)/2] = 0
-2sin[4x/2]sin[-2x/2] = 0
-2sin(2x)sin(-x) = 0
sin(2x)sin(x) = 0

Now we have a product of two factors that equals zero. This means that either sin(2x) = 0 or sin(x) = 0. So our solutions are:

a) sin(2x) = 0
2x = nπ where n is an integer
x = nπ/2

b) sin(x) = 0
x = nπ where n is an integer

Therefore, the solutions to cos^2(3x) - cos(3x)cos(5x) = 0 are x = (2n + 1)π/6, x = nπ/2, and x = nπ where n is an integer.