1) 3sin (π/2-π) - √3=02) 3ctg (π/2+x) - √3=0

Найдем корни уравнений.

1) 3 * sin (π/2 - π) - √3 = 0;

3 * sin (-π/2) - √3 = 0;

3 * sin (-π/2) = √3;

-3 * sin (π/2) = √3;

sin (π/2) = √3 / (-3);

sin (π/2) = - √3/3;

Уравнение не имеет корней.

2) 3 * ctg (π/2 + x) - √3 = 0;

3 * ctg (π/2 + x) = √3;

ctg (π/2 + x) = √3/3;

π/2 + x = arcctg (√3/3) + π * n, n ∈ Z;

π/2 + x = π / 6 + π * n, n ∈ Z;

x = π / 6 - π/2 + π * n, n ∈ Z;

x = π / 6 - 3 * π/6 + π * n, n ∈ Z;

x = - 2 * π/6 + π * n, n ∈ Z;

x = - π/3 + π * n, n ∈ Z;

Ответ: x = - π/3 + π * n, n ∈ Z.