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