:1) 2cos(п/3-2x)= -1 2) √2cos(x/3+п/4)=1 3) 2cos(п/6+3x)=√3

Vasya1337ez Vasya1337ez    1   01.10.2019 01:10    0

Ответы
npapriko npapriko  21.08.2020 19:27

1) 2cos(п/3-2x)= -1

cos(2х -п/3)= -1 /2

2х - π/3 = +-arcCos(-1/2) +2πk , k ∈Z

2x - π/3 = +-2π/3 + 2πk , k ∈Z

2x = π/3 +-2π/3 + 2πk , k ∈Z

x = π/6 +-π/3 + πk , k ∈Z

2) √2cos(x/3+п/4)=1

Cos(x/3 + π/4 ) = 1/√2

x/3 + π/4 = +-arcCos1/√2 + 2πk , k ∈Z

x/3 + π/4 = +-π/4 + 2πk , k ∈Z

x/3 = -π/4 +-π/4 + 2πk , k ∈Z

x = -3π/4 +-3π/4 + 6πk , k ∈Z

3) 2cos(п/6+3x)=√3

Cos(π/6+3x) = √3/2

π/6 + 3x = +-arcCos√3/2 + 2πk , k ∈Z

3x = -π/6 +-π/6 + 2πk , k ∈ Z

x = -π/18 +-π/18 + 2πk/3 , k∈Z

ПОКАЗАТЬ ОТВЕТЫ
Другие вопросы по теме Алгебра