1) arccos (-1) - arccos 1/2 - 3 arccos ( - √3/2) 2) 2 сos t=1 3) - 2cos t=0 4) cos (arcctg √3) 5) arcctg (ctg 2 П/3) 6) ctg x=-0,5 7) ctg x=0

1)

arccos (-1) - arccos (1/2) - 3arccos (-√3/2) = - π - π/3 - 3 * 5π/6 = - 6π/6 - 2π/6 - 15π/6 = - 23π/6.

2)

2 сost = 1; сost = 1/2; t = ±π/3 + 2πk, k ∈ Z.

3)

-2cost = 0; cost = 0; x = π/2 + πk, k ∈ Z.

4)

cos (arcctg√3) = cos (π/6) = √3/2.

5)

arcctg (ctg (2π/3)) = 2π/3.

6)

ctgx = - 0,5; x = - arcctg (0,5) + πk, k ∈ Z.

7)

ctgx = 0; x = π/2 + πk, k ∈ Z.

Ответ:

1) - 23π/6; 2) ±π/3 + 2πk, k ∈ Z; 3) π/2 + πk, k ∈ Z; 4) √3/2; 5) 2π/3; 6) - arcctg (0,5) + πk, k ∈ Z; 7) π/2 + πk, k ∈ Z.