Question

4sin3x+sin5x-2sinx cos2x=0

Answer 1

Ответ:

$4 \sin(3x) + \sin(5x)$

Answer 2

To solve the trigonometric equation 4sin(3x) + sin(5x) - 2sin(x)cos(2x) = 0, we will use trigonometric identities and algebraic manipulation. Let's go step by step:

Step 1: Apply the double angle identity for cosine, sin(2x) = 2sin(x)cos(x): 4sin(3x) + sin(5x) - 2sin(x)(2cos^2(x) - 1) = 0

Step 2: Expand the expression: 4sin(3x) + sin(5x) - 4sin(x)cos^2(x) + 2sin(x) = 0

Step 3: Apply the multiple angle identity for sine, sin(3x) = 3sin(x) - 4sin^3(x), and sin(5x) = 5sin(x) - 20sin^3(x): 4(3sin(x) - 4sin^3(x)) + (5sin(x) - 20sin^3(x)) - 4sin(x)cos^2(x) + 2sin(x) = 0

Step 4: Distribute and combine like terms: 12sin(x) - 16sin^3(x) + 5sin(x) - 20sin^3(x) - 4sin(x)cos^2(x) + 2sin(x) = 0

Step 5: Combine the sine terms: 17sin(x) - 36sin^3(x) - 4sin(x)cos^2(x) + 2sin(x) = 0

Step 6: Factor out sin(x): sin(x)(17 - 36sin^2(x) - 4cos^2(x) + 2) = 0

Step 7: Apply the Pythagorean identity, cos^2(x) = 1 - sin^2(x): sin(x)(17 - 36sin^2(x) - 4(1 - sin^2(x)) + 2) = 0

Step 8: Simplify the expression: sin(x)(17 - 36sin^2(x) - 4 + 4sin^2(x) + 2) = 0

Step 9: Combine like terms: sin(x)(15 - 32sin^2(x)) = 0

Now, we have two possibilities for the equation to be true:

1. sin(x) = 0
2. 15 - 32sin^2(x) = 0

Solving for sin(x) = 0, we get x = 0, π, 2π, 3π, ...

Solving for 15 - 32sin^2(x) = 0, we get:

32sin^2(x) = 15

sin^2(x) = 15/32

Taking the square root of both sides:

sin(x) = ±√(15/32)

sin(x) = ±√(15)/√(32)

sin(x) ≈ ±0.608

Since sine values fall between -1 and 1, the equation sin(x) ≈ 0.608 has solutions within that range.

To summarize, the solutions to the equation 4sin(3x) + sin(5x) - 2sin(x)cos(2x) = 0 are:

1. x = 0, π, 2π, 3π, ... (where sin(x) = 0)
2. x ≈ arcsin(√(15)/√(32)) + 2πk or x ≈ π - arcsin(√(15)/√(32)) + 2πk (where k is an integer)