Question

2sin-√3=0
Cos2x=√2/2
Sin+1/√2=0

Answer 1

1)2sinx-√3=0, 2sinx=√3, sinx=(√3)/2
   sin(π/3)=(√3)/2 , sin(2π/3)=(√3/2)
   x1=π/3 +2kπ, k∈Z
   x2=2π/3 + 2kπ, k∈Z
2)cos2x=(√2)/2
   cos(π/4)=(√2)/2 , cos(7π/4)=(√2)/2
   2x=π/4+2kπ, x1=π/8+kπ , k∈Z
   2x=7π/4+2kπ, x2=7π/8 +kπ, k∈Z
3)sinx+1/√2=0, sinx=-1/√2, sinx=-√2/2
  sin(5π/4)=-√2/2 , sin(7π/4)=-√2/2
  x1=5π/4+2kπ,k∈Z
  x2=7π/4+2kπ,k∈Z