Question

Sin3xctgx=0 пжжжжжжжжжжжжжжж

Answer 1

   1. Область допустимых значений переменной:

sinx ≠ 0;x ≠ πk, k ∈ Z.

   2. Приравниваем каждый из множителей к нулю:

sin(3x) * ctgx = 0;[sin(3x) = 0;

[ctgx = 0.

   3. Функции синус и косинус периодические с периодами 2π:

[sin(3x) = 0;

[ctgx = 0;[sin(3x) = 0;

[cosx/sinx = 0;[sin(3x) = 0;

[cosx = 0;[3x = πk, k ∈ Z;

[x = π/2 + πk, k ∈ Z;[x = πk/3, k ∈ Z;

[x = π/2 + πk, k ∈ Z.

   4. С учетом области допустимых значений переменной получим:

      x = π/2 + πk; π/2 ± π/6 + πk, k ∈ Z.

   Ответ: π/2 + πk; π/2 ± π/6 + πk, k ∈ Z.

Answer 2

To solve the equation sin(3x)ctg(x) = 0, we can break it down into two separate equations and solve each one individually.

First, let's consider the equation sin(3x) = 0. This equation implies that the sine of 3x is equal to zero. The sine function equals zero at certain specific angles. In this case, we can write:

3x = nπ, where n is an integer.

To solve for x, we divide both sides of the equation by 3:

x = nπ/3, where n is an integer.

Now, let's consider the equation ctg(x) = 0. The cotangent function equals zero at certain specific angles. In this case, we can write:

cot(x) = 0

The cotangent function is the reciprocal of the tangent function. It equals zero when the tangent of the angle is infinity or when the angle itself is an odd multiple of π/2 (90 degrees). Thus, we have:

x = (2n + 1)π/2, where n is an integer.

So, the solutions to the equation sin(3x)ctg(x) = 0 are given by the values of x that satisfy both equations. Therefore, the solutions are:

x = nπ/3 or x = (2n + 1)π/2, where n is an integer.