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.