Question

помогите срочно ......

Answer 1

Решение
1.
а)  sin(4π/3) = - √3/2
б)  cos210° ≈ 0,2079
в)  ctg(- 135°) = - ctg135° = - (-1) = 1
2.  
а)  sinx = - √2/2
x = (-1)^n arcsin(-√2/2) + πk, k∈Z
x = (-1)^(n+1) arcsin(√2/2) + πk, k∈Z
x = (-1)^(n+1) (π/4) + πk, k∈Z
б)  cosx = √3/2
x = (+ -)arccos(√3/2) + 2πn, n∈Z
x = (+ -)arccos(π/6) + 2πn, n∈Z
3.
а)   cos0 * sinπ/2 * c0s5π = 1 * 1*(-1) = - 1
б)  tg60 ° - ctg60° = √3 - 1/√3 = 2√3/3
4, 
2sinx - √3 = 0
sinx = √3/2
x = (-1)^n arcsin(√3/2) + πk, k∈Z
x = (-1)^n(π/3) + πk, k∈Z
5.  
4tg²x - 12tgx + 5 = 0
tgx = t
4t² - 12t + 5 = 0
D = 144 - 4*4*5 = 64
t₁ = (12 - 8)/8
t₁ = 1/2
t₂ = (12 + 8)/8
t₂ = 5/2
1)  tgx = 1/2
x₁ = arctg(1/2) + πk, k∈Z
2)  tgx = 5/2
x₂ = arctg(5/2) + πn, n∈Z
Ответ: x₁ = arctg(1/2) + πk, k∈Z ; x₂ = arctg(5/2) + πn, n∈Z