Question

1)cos(п/2-x)-sin3x+sin5x=02)cos2x+cos4x+cos(п-3x)=0

Answer 1

Смотри фото решение на листе

Answer 2

Let's solve each equation one by one.

1. cos(π/2 - x) - sin(3x) + sin(5x) = 0

To simplify this equation, we can make use of the trigonometric identities:

cos(π/2 - x) = sin(x)

Substituting this into the equation, we get:

sin(x) - sin(3x) + sin(5x) = 0

To solve this equation, we'll apply the sum-to-product identities:

sin(a) - sin(b) = 2 * cos((a + b)/2) * sin((a - b)/2)

Using this identity, the equation becomes:

2 * cos((3x + x)/2) * sin((3x - x)/2) + sin(5x) = 0

Simplifying further:

2 * cos(2x) * sin(x) + sin(5x) = 0

Now, let's use another trigonometric identity:

sin(5x) = -sin(x)

Substituting this back into the equation, we have:

2 * cos(2x) * sin(x) - sin(x) = 0

Factoring out sin(x):

sin(x) * (2 * cos(2x) - 1) = 0

This equation will be satisfied if either sin(x) = 0 or 2 * cos(2x) - 1 = 0.

If sin(x) = 0, then x can be any multiple of π (x = nπ, where n is an integer).

If 2 * cos(2x) - 1 = 0, we can solve for cos(2x):

2 * cos(2x) - 1 = 0 2 * cos(2x) = 1 cos(2x) = 1/2

Using the inverse cosine function:

2x = ±π/3 + 2πn, where n is an integer

Simplifying:

x = ±π/6 + πn, where n is an integer

So the solutions to the equation are x = nπ (where n is an integer) and x = ±π/6 + πn (where n is an integer).

2. cos(2x) + cos(4x) + cos(π - 3x) = 0

Using the identity cos(π - x) = -cos(x), the equation becomes:

cos(2x) + cos(4x) - cos(3x) = 0

Now, we'll use the sum-to-product identities to simplify the equation:

cos(a) + cos(b) = 2 * cos((a + b)/2) * cos((a - b)/2)

cos(a) - cos(b) = -2 * sin((a + b)/2) * sin((a - b)/2)

Applying these identities, the equation becomes:

2 * cos(3x) * cos(x) - 2 * sin(3x) * sin(x) - cos(3x) = 0

Simplifying further:

cos(3x) * (2 * cos(x) - 1) - 2 * sin(3x) * sin(x) = 0

Using the identity sin(3x) = 3 * sin(x) - 4 * sin^3(x):

cos(3x) * (2 * cos(x) - 1) - 2 * (3 * sin(x) - 4 * sin^3(x)) * sin(x) = 0

Expanding and simplifying:

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