Question

Срочно надо❤️
Sosososos
22 балла отдаю

Answer 1

Ответ:

Объяснение:

a) 2cos²x - cosx-1 = 0

cosx = y

2y²-y-1 = 0

D = 1+8 = 9

y ₁ ₂ = (1±3)/4

y₁ = 1

y₂ = -1/2

cosx = 1

x = 2πn n∈Z

cosx = -1/2

x = 2π/3 + 2πn

x = 4π/3 + 2πn

n∈Z

б) 3sin²x+sinx*cosx-2cos²x = 0

разделим все на cos²x

3tg²x+tgx-2 = 0

tgx = y

3y²+y-2 = 0

D = 1+24 = 5²

sinx ≥√3/2

x∈ [π/3 + 2πn; 2π/3+2πn]

y ₁ ₂ = (-1 ±5)/6

y₁ = -1

y₂ = 2/3

tgx = 1  x = -π/4 + πn

tgx = 2/3

x = arctg(2/3) + πn

в) аналогично делим на cos x

3tgx = 2

tgx = 2/3

x = arctg(2/3) + πn

cosx <-$\sqrt{3} /2$

x∈ (5π/6 + 2πn; 7π/6 + 2πn)