1 - 2 sin (θ) = cos (2θ)

Cos (2 Ф) = cos² (Ф) - sin² (Ф)

sin² (Ф) + cos² (Ф) = 1 ⇒cos² (Ф) = 1-sin² (Ф)

Then:

1-2sin (Ф) = cos (2 Ф)

1-2sin (Ф) = cos² (Ф) - sin² (Ф)

1-2sin (Ф) = (1-sin² (Ф)) - sin² (Ф)

1-2sin (Ф) = 1-2sin² (Ф)

2sin² (Ф) - sin (Ф) + 1-1=0

2sin² (Ф) - 2sin (Ф) = 0

2sin (Ф) (sin (Ф) - 1) = 0

We have to solve two equations:

1)

2sin (Ф) = 0

sin (Ф) = 0/2

sin (Ф) = 0

Ф=sin⁻¹ 0=kπ (1) K∈Z

2)

sin (Ф) - 1=0

sin (Ф) = 1

Ф=sin⁻¹ 1=π/2 + 2Kπ (2) K∈Z

Solution=solution₁ + solution₂

Answer: kπ U π/2+2kπ; K = ( ... - 3.-2,-1,0,1,2,3, ...)