Question

cos ( 3x+ п/6) <- √2/2 пожалуйста помогите !!!!!!​ очень срочно!!!! sos ​

Answer 1

Ответ:

cos (3x + π/6) < -√2/2

Замена (3x + π/6) = t

cost < -√2/2

arccos(-√2/2) + 2πk < t < 2π - arccos(-√2/2) + 2πk , k € Z

(π - arccos(√2/2)) + 2πk < t < 2π - (π - arccos(√2/2)) + 2πk , k € Z

3π/4 + 2πk < t < 2π - 3π/4 + 2πk , k € Z

3π/4 + 2πk < t < 5π/4 + 2πk , k € Z

3π/4 + 2πk < 3x + π/6 < 5π/4 + 2πk , k € Z

3π/4 - π/6 + 2πk < 3x < 5π/4 - π/6 + 2πk , k € Z

9π/12 - 2π/12 + 2πk < 3x < 15π/12 - 2π/12 + 2πk , k € Z

7π/12 + 2πk < 3x < 13π/12 + 2πk , k € Z

7π/36 + 2πk/3 < x < 13π/36 + 2πk/3 , k € Z