log2(2-cosx)=1+2log2(-sinx)

Loq2 (2 - cosx )=Loq2 2 +2Loq2 (-sinx) ,  ОДЗ : sinx < 0  
2 -cosx =2sin²x ;
2 -cosx  =2(1-cos²x) ;
2cosx(cosx -1/2) =0 ;
[ cosx =0 ;cosx= 1/2,
cosx=0⇒sinx =+ -1⇒sinx = -1⇔x = -π/2 +2π*k ,k∈Z.
cosx= 1/2⇒x = -π/3+2π*k  ,k∈Z.

ответ : π/2 +2π*k     ;   -π/3+2π*k  ,k∈Z.