а) 2sinx + 1 = 0 ; б) x € [-2π ; 0]
а) 2sinx = -1
sinx = -0,5
x = -π/6 + 2πk ; x = -5π/6 + 2πk ; k € Z
б) -2π =< -π/6 + 2πk =< 0
-11π/6 =< 2πk =< π/6
-11/12 =< k =< 1/12
k € Z ; k = 0
x = -π/6
-2π =< -5π/6 + 2πk =< 0
-7π/6 =< 2πk =< 5π/6
-7/12 =< k =< 5/12
k € Z ; k = 0
x = -5π/6
ответ : а) x = -π/6 + 2πk ; x = -5π/6 + 2πk ; k € Z ; б) -5π/6 ; -π/6
