Question

Sin2x-sin8x-cos10x=1

Answer 1

Ответ:

Пошаговое объяснение:

sin2x-sin8x = 2*(-sin3x)* cos5x

2*(-sin3x)* cos5x - cos^2 (5x) + sin^2 (5x) - sin^2 (5x) - cos^2 (5x) = 0

2*(-sin3x)* cos5x - 2cos^2 (5x) = 0  

sin3x * cos5x + cos^2 (5x) = 0

cos5x * (sin3x + cos5x) = 0

cos5x * (sin3x + sin(pi/2 - 5x)) = 0

cos5x * 2sin(pi/4 - x) * cos(4x - pi/4) = 0

cos5x = 0

5x = pi/2 + pin

x1 = pi/10 + pin/5, n ∈ Z

sin(pi/4 - x) = 0

pi/4 - x = pik

x2 = pi/4 - pik, k ∈ Z

cos(pi/4 + 4x) = 0

4x - pi/4 = pi/2 + pim

4x = 3pi/4 + pim

x3 = 3pi/16 + pim/4, m ∈ Z

Answer 2

To solve the equation sin(2x) - sin(8x) - cos(10x) = 1, we'll break it down into simpler parts and solve each component individually.

Let's start by rewriting the equation:

sin(2x) - sin(8x) - cos(10x) = 1

To solve trigonometric equations like this, it's helpful to use trigonometric identities and properties. In this case, we'll use the sum-to-product and double angle identities.

Step 1: Simplifying sin(2x)

We'll use the double angle identity sin(2θ) = 2sin(θ)cos(θ) to simplify sin(2x):

sin(2x) = 2sin(x)cos(x)

The equation becomes:

2sin(x)cos(x) - sin(8x) - cos(10x) = 1

Step 2: Simplifying sin(8x)

We'll use the sum-to-product identity sin(A) - sin(B) = 2cos((A + B)/2)sin((A - B)/2) to simplify sin(8x):

sin(8x) = sin(10x - 2x) = 2cos(6x)sin(2x)

The equation becomes:

2sin(x)cos(x) - 2cos(6x)sin(2x) - cos(10x) = 1

Step 3: Simplifying cos(10x)

We'll use the double angle identity cos(2θ) = 1 - 2sin²(θ) to simplify cos(10x):

cos(10x) = cos(8x + 2x) = cos²(8x) - sin²(8x) = (1 - 2sin²(4x))(1 - 2sin²(4x)) = 1 - 4sin²(4x) + 4sin⁴(4x)

The equation becomes:

2sin(x)cos(x) - 2cos(6x)sin(2x) - (1 - 4sin²(4x) + 4sin⁴(4x)) = 1

Simplifying further:

2sin(x)cos(x) - 2cos(6x)sin(2x) + 4sin²(4x) - 4sin⁴(4x) + 1 = 1

2sin(x)cos(x) - 2cos(6x)sin(2x) + 4sin²(4x) - 4sin⁴(4x) = 0

Step 4: Rearranging terms

Rearranging the terms in descending powers of sin:

-4sin⁴(4x) + 4sin²(4x) + 2sin(x)cos(x) - 2cos(6x)sin(2x) = 0

Step 5: Factoring out sin²(4x)

Factoring out sin²(4x):

sin²(4x)(-4sin²(4x) + 4) + 2sin(x)cos(x) - 2cos(6x)sin(2x) = 0

Step 6: Simplifying further

Using the Pythagorean identity sin²(θ) + cos²(θ) = 1, we can rewrite -4sin²(4x) + 4 as 4cos²(4x):

sin²(4x)(4cos²(4x)) + 2sin(x)cos(x