cos (2Ф)=cos²(Ф)-sin²(Ф)
sin²(Ф)+cos²(Ф)=1 ⇒cos²(Ф)=1-sin²(Ф)
Then:
1-2sin(Ф)=cos(2Ф)
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,...)
