Question

решить неравенства: 1) 2 сos ( 2x-π/6)≤1; 2) √2 sin(π/4+x/2)≥1:​

Answer 1

1)  сos ( 2x-(π/6))≤1/2

(π/3)+2πn≤( 2x-(π/6))≤(5π/3)+2πn, n∈Z

(π/3)+(π/6)+2πn≤ 2x-(π/6)+(π/6)≤(5π/3)+(π/6)+2πn, n∈Z

(3π/6)+2πn≤ 2x≤(11π/6)+2πn, n∈Z

(3π/12)+πn≤ x≤(11π/12)+πn, n∈Z

(π/4)+πn≤ x≤(11π/12)+πn, n∈Z -   о т в е т

2)

sin((π/4)+(x/2))≥1/√2

(π/4)+2πk≤((π/4)+(x/2))≤(3π/4)+2πk, k∈Z

(π/4)-(π/4)+2πk≤(x/2)≤(3π/4)-(π/4)+2πk, k∈Z

2πk≤(x/2)≤(2π/4)+2πk, k∈Z

4πk≤x≤π+4πk, k∈Z  -   о т в е т

Answer 2

Ответ:

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Объяснение: