Verified answer

Sin(π/2 + x) - Cos(π - x) + 1 = 0

Sin(a+b) = Sina*Cosb + Cosa*Sinb

Cos(a-b) = Cosa*Cosb + Sina*Sinb

1)Sin(π/2 + x) = Sin(π/2)*Cosx + Cos(π/2)*Sinx = Cosx

 Cos(π - x) = Cosπ*Cosx + Sinπ*Sinx = -Cosx

 Cosx - (-Cosx) + 1 = 0

 2Cosx + 1 = 0

 Cosx = -1/2

 x = ±2π/3 + 2πn, n∈Z

Sina - Sinb = 2Sin((a-b)/2)*Cos((a+b)/2)

2) Sin(7x) = Sin(5x)

   Sin(7x) - Sin(5x) = 0

   2Sinx *Cos(6x) = 0

   Sinx = 0 => x = πn, n∈Z

   Cos(6x) = 0 => 6x = π/2 + 2πn, n∈Z

                              x = π/12 + πn/3, n∈Z