Question

1) 2cos x= √3;
2) sin x= 1/2;
3) cos 2x= -1

Answer 1

1) 2cox = √3
    cosx = √3/2
    x = +- arccos(√3/2) + 2πk, k ∈ Z
    x = +- π/6 + 2πk
2)sin x = 1/2
   x = (-1)^k arcsin(1/2) + πk, k ∈ Z
   x = (-1)^k π/6 + πk
3)cos2x = -1
   2x = π + 2πk, k ∈ Z 
   x = π/2 + πk