Question

1)      3tg*x/2= √3

2)      cos 2/3*x = 1/2

3)      sinx ≥ √3/2

4)      sin 3π/√х = - √3/2

Answer 1

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