1) 3tg*x/2= √3 2) cos 2/3*x = 1/2 3) sinx ≥ √3/2 4) sin 3π/√х = - √3/2

Question

Алгебра: 1) 3tg*x/2= √3 2) cos 2/3*x = 1/2 3) sinx ≥ √3/2 4) sin 3π/√х = - √3/2

neznaika167 · Answer

1) 3 tg x/2=√3

tgx/2=√3/3

x/2= π/6+πn

x= π/3+2πn

2) cos 2/3*x = 1/2

2/3*x= ± π/3

2x=±π+6πn

x=±π/2= 3 πn

3) sinx ≥ √3/2

x∈ [ π/3+ 2πn; 2π/3 + 2πn ] n∈ Z

1) 3tgx/2= √3 2) cos 2/3x = 1/2 3) sinx ≥ √3/2 4) sin 3π/√х = - √3/2