Question

до завтра!помогите,буду благодарна

1. Решите уравнение а) $sqrt{3}$ ctg $frac{x}{2}$ = 3
б) sin ($frac{ pi }{6}$ - 2x) = 0

Answer 1

a) √3ctg(x/2) = 3
ctg(x/2) = 3/√3
ctg(x/2) = √3
x/2 = π/6 + πk, k€Z
x = π/12 + πk/2, k€Z.

b) sin(π/6 - 2x) = 0
-sin(2x - π/6) = 0
sin(2x - π/6) = 0
-π/6 + 2x = πk, k€Z
2x = π/6 + πk, k€Z
x = π/12 + πk/2, k€Z.