А) решаем как неполное квадратное: Cos x(2Cos x + √3) = 0 Cos x = 0 или 2Сos x + √3 = 0 x = π/2 + πk , k ∈Z 2Cos x = -√3 Cos x = -√3/2 x = +-arcCos(-√3/2) + 2πn , n ∈Z x = +- 5π/6 + 2πn , n ∈Z б) Решаем как неполное квадратное: tg x(√3 tgx -3) = 0 tg x = 0 или √3 tgx -3 = 0 x = πn, n ∈Z √3 tg x = 3 tg x = 3/√3 = √3 x = π/3 + πk , k ∈Z в) √3 Sin x + Cos x = 0 |:2 √3/2 Sin x + 1/2Cos x = 0 Sin π/3 Sin x + Cos π/3 Cos x = 0 Cos(π/3 -x) = 0 π/3 - х = π/2 + πk , k ∈ Z -x = -π/3 + π/2 + πk, k ∈Z -х = π/6 + πk , k ∈Z x = -π/6 - πk, k ∈Z г) Sin 2x + Cos 2x = √2 | : √2 1/√2 ·Sin 2x + 1/√2·Cos 2x =1 Sinπ/4 Sin 2x + Cos π/4 Cos 2x = 1 Cos(π/4 - 2x) = 1 π/4 - 2x = 2πk , k ∈Z -2x = -π/4 + 2πk , k ∈Z х = π/8 - πk , k ∈Z
Answers & Comments
Verified answer
А) решаем как неполное квадратное:Cos x(2Cos x + √3) = 0
Cos x = 0 или 2Сos x + √3 = 0
x = π/2 + πk , k ∈Z 2Cos x = -√3
Cos x = -√3/2
x = +-arcCos(-√3/2) + 2πn , n ∈Z
x = +- 5π/6 + 2πn , n ∈Z
б) Решаем как неполное квадратное:
tg x(√3 tgx -3) = 0
tg x = 0 или √3 tgx -3 = 0
x = πn, n ∈Z √3 tg x = 3
tg x = 3/√3 = √3
x = π/3 + πk , k ∈Z
в) √3 Sin x + Cos x = 0 |:2
√3/2 Sin x + 1/2Cos x = 0
Sin π/3 Sin x + Cos π/3 Cos x = 0
Cos(π/3 -x) = 0
π/3 - х = π/2 + πk , k ∈ Z
-x = -π/3 + π/2 + πk, k ∈Z
-х = π/6 + πk , k ∈Z
x = -π/6 - πk, k ∈Z
г) Sin 2x + Cos 2x = √2 | : √2
1/√2 ·Sin 2x + 1/√2·Cos 2x =1
Sinπ/4 Sin 2x + Cos π/4 Cos 2x = 1
Cos(π/4 - 2x) = 1
π/4 - 2x = 2πk , k ∈Z
-2x = -π/4 + 2πk , k ∈Z
х = π/8 - πk , k ∈Z