Question

помогите срочно даю 100 баллов

Answer 1

1. а)  cos4x = 0

4x = (π/2) + πn

x = (π/8) + (πn/4) , n ∈ Z

ОТВЕТ: (π/8) + (πn/4) , n ∈ Z

б)  tg2x = - 1

2x = (-π/4) + πn

x = (-π/8) + (πn/2) , n ∈ Z

ОТВЕТ: (-π/8) + (πn/2) , n ∈ Z

2. 2sin(3x - (π/6)) + √3 = 0

2sin(3x - (π/6)) = - √3

sin(3x - (π/6)) = - √3/2

[ 3x - (π/6) = - π/3 + 2πk  ⇔  3x = (-π/3) + 2πk  ⇔  x = (-π/9) + (2πk/3)

[ 3x - (π/6) = - 2π/3 + 2πk  ⇔  3x = (-π/2) + 2πk  ⇔  x = (-π/6) + (2πn/3) , k ∈ Z

ОТВЕТ: (-π/9) + (2πk/3) , (-π/6) + (2πk/3) , k ∈ Z

3. 2cos²x - 3√3cosx + 3 = 0

Пусть cosx = a, a ∈ [ - 1 ; 1 ] , тогда

2a² - 3√3a + 3 = 0

D = (3√3)² - 4•2•3 = 27 - 24 = 3

a = (3√3 - √3)/4 = 2√3/4 = √3/2

a = (3√3 + √3)/4 = 4√3/4 = √3  ∉ [ - 1 ; 1 ]

a = √3/2 ⇔ cosx = √3/2 ⇔ x = ± (π/6) + 2πm, m ∈ Z

ОТВЕТ: ± (π/6) + 2πm , m ∈ Z