Question

Помогите решить уравнение: 2sinx+1=0 на интервале [-2п;0]

Answer 1

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