Ответы на вопрос:
Решение 1) √2sinx-1≥0 sinx ≥ 1/√2 arcsin(1/√2) + 2πn ≤ x ≤ π - arcsin(1/√2) + 2πn, n∈z π/4 + 2πn ≤ x ≤ π - π/4 + 2πn, n∈z π/4 + 2πn ≤ x ≤ 3π/4 + 2πn, n∈z 2) 2cos(2xπ/6)> √3 cos(2xπ/6) > √3/2 - arccos(√3/2) + 2πk < 2xπ/6 < arccos(√3/2) + 2πk, k∈z - π/6 + 2πk < 2xπ/6 < π/6 + 2πk, k∈z - 1/2 + 6k < x < 1/2 + 6k, k∈z - 1/2 < x < 1/2
ответ:
x во 2+8=6x
x во 2+8-6x=0
x во 2-6x+8=0
x во 2-2x-4x+8=0
x(x-2)-4(x-2)=0
(x-2)(x-4)=0
x-2=0
x-4=0
x1=2
x2=4
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