Ответ(ы) на вопрос:
Гость 1) √3tg²(3x) - 3tg(3x) = 0
tg(3x)(√3tg(3x) - 3) = 0
a) tg3x = 0
3x = πk, k∈Z
x = πk/3, k∈Z
b) √3tg(3x) - 3 = 0
tg(3x) = 3/√3
3x = arctg(3/√3) + πn, n∈Z
3x = π/3 + πn, n∈z
x = π/9 + πn/3, n∈Z
2) 4sin²(2x + π/3) - 1 = 0
sin²(2x + π/3) = 1/4
a) sin(2x + π/3) = - 1/2
2x + π/3 = (-1)^(n)*arcsin(-1/2) + 2πk, k∈Z
2x + π/3 = (-1)^(n+1)*arcsin(1/2) + 2πk, k∈Z
2x + π/3 = (-1)^(n+1)*(π/6) + 2πk, k∈Z
2x = (-1)^(n+1)*(π/6) - π/3 + 2πk, k∈Z
x = (-1)^(n+1)*(π/12) - π/6 + πk, k∈Z
b) sin(2x + π/3) = 1/2
2x + π/3 = (-1)^(n)*arcsin(1/2) + 2πk, k∈Z
2x + π/3 = (-1)^(n)*(π/6) + 2πk, k∈Z
2x = (-1)^(n)*(π/6) - π/3 + 2πk, k∈Z
x = (-1)^(n)*(π/12) - π/6 + πk, k∈Z
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