Question

Sin (4x) + cos (6x) = 0

Answer 1

Sin (4x) + cos (6x) = 0;
Sin (4x) = -cos (6x);
Sin (4x) = sin (3π/2+6x);
Sin (4x) - sin (3π/2+6x) = 0;
2sin ((3π+4x)/4) * cos ((3π+20x)/4)=0;
sin ((3π+4x)/4) = 0 или cos ((3π+20x)/4)=0;
Решаем первое:
sin ((3π+4x)/4) = 0;
(3π+4x)/4 = πn, n∈Z;
4x = 4πn - 3π, n∈Z;
x = πn - 3π/4, n∈Z - наш первый ответ.

Решаем второе:
cos ((3π+20x)/4)=0;
(3π+20x)/4 = π/2 +πn, n∈Z;
20x = -π + 4πn, n∈Z;
x = -π/20 +πn/5, n∈Z - наш второй ответ.

Answer 2

To solve the equation sin(4x) + cos(6x) = 0, we can use the trigonometric identity:

sin^2(x) + cos^2(x) = 1

We can rewrite sin(4x) and cos(6x) in terms of sin(x) and cos(x) using the following identities:

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

Substituting these identities into the original equation, we get:

2sin(2x)cos(2x) + cos(2x) - 2sin^2(2x)cos(2x) - 2sin(2x)cos(2x) = 0

Simplifying this equation, we get:

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

This equation is satisfied if either cos(2x) = 0 or 1 - 2sin^2(2x) - 4sin(2x) = 0.

If cos(2x) = 0, then 2x = π/2 + kπ for some integer k. This gives x = π/4 + kπ/2.

If 1 - 2sin^2(2x) - 4sin(2x) = 0, then we can use the substitution y = sin(2x) to get:

1 - 2y^2 - 4y = 0

Solving for y, we get:

y = (-b ± √(b^2 - 4ac)) / 2a

where a = -2, b = -4, and c = 1. Substituting these values, we get:

y = (2 ± √(20)) / (-4) = (-1 ± √5) / 2

Substituting y = sin(2x), we get:

sin(2x) = (-1 ± √5) / 2

This equation has two solutions:

2x = arcsin[(-1 + √5) / 2] + kπ or 2x = arcsin[(-1 - √5) / 2] + kπ

Solving for x, we get:

x = (arcsin[(-1 + √5) / 2] + kπ) / 2 or x = (arcsin[(-1 - √5) / 2] + kπ) / 2

Therefore, the solutions to the equation sin(4x) + cos(6x) = 0 are:

x = π/4 + kπ/2 or x = (arcsin[(-1 + √5) / 2] + kπ) / 2 or x = (arcsin[(-1 - √5) / 2] + kπ) / 2, where k is any integer.