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Valeriyamiers

12 авг. 2013 г., 4:41:13 (10 лет назад)

1)  sin (t+П/5) =√2/2
t +π/5 = (-1)^n*arcsin(√2/2) + πn, n∈Z
t +π/5 = (-1)^n*(π/4) + πn, n∈Z
t = (-1)^n*(π/4) - π/5 + πn, n∈Z
2) сos (2t +П/4)=0
2t + π/4 =  π/2 + πk, k∈Z
2t  =  π/2 - π/4 + πk, k∈Z
2t  = π/4 + πk, k∈Z
t  = π/8 + πk/2, k∈Z
3) tg(t/2- П/2) = - √3
 - tg( π/2- t/2) = - √3 
- ctg(t/2) = - √3 
 ctg(t/2) =  √3
t/2 = arctg(√3) + πn, n∈Z
t/2 = π/3 + πn, n∈Z
t = 2π/3 + 2πn, n∈Z
4) сos^ 2(2t + π/6) = 1/2 
 a)   сos(2t + π/6) = -√2/2
2t + π/6 = (+ -)*arccos(-√2/2) + 2πk, k∈Z
 2t + π/6 = (+ -)*(π - π/4) + 2πk, k∈Z
 2t + π/6 = (+ -)*(3π/4) + 2πk, k∈Z
 2t  = (+ -)*(3π/4)  - π/6 + 2πk, k∈Z
t1  = (+ -)*(3π/8)  - π/12 + πk, k∈Z
b)   сos(2t + π/6) = √2/2
2t + π/6 = (+ -)*arccos(√2/2) + 2πk, k∈Z
2t + π/6 = (+ -)*(π/4) + 2πk, k∈Z
2t = (+ -)*(π/4) - π/6 + 2πk, k∈Z
t2 = (+ -)*(π/8) - π/12 + πk, k∈Z
5) ctg^ 2(2t - П/2)= 1/3
a)  ctg(2t - П/2)= - √3/3
2t - π/2 = arcctg(-√3/3) + πn, n∈Z
2t - π/2 = 2π/3 + πn, n∈Z
2t  = 2π/3 + π/2+ πn, n∈Z
2t  = 7π/6 + πn, n∈Z
t1  = 7π/12 + πn/2, n∈Z
b)  ctg(2t - П/2)=  √3/3
2t - π/2 = arcctg(√3/3) + πn, n∈Z
2t - π/2 = π/3 + πk, n∈Z
2t  = π/3 + π/2+ πk, n∈Z
2t  = 5π/6 + πk, n∈Z
t2  =5π/12 + πk/2, n∈Z
6) tg ^2 (3t+П/2)=1/3
 a)  tg  (3t+π/2) = - √3/3
-ctg(3t)= -√3/3
ctg(3t)= √3/3
3t  = arcctg(√3/3) + πn, n∈Z
3t  = π/3 + πk, n∈Z
t1  = π/9 + πk/3, n∈Z
b)  tg  (3t+π/2) =  √3/3
ctg(3t)= - √3/3
3t  = arcctg(-√3/3) + πn, n∈Z
3t  = 2π/3 + πn, n∈Z
t  = 2π/9 + πn/3, n∈Z
7) 3 cos ^2t - 5 cos t = 0
cost(3cost - 5) = 0
a)  cost = 0
 t =  π/2 + πn, n∈Z
b)  3cost - 5 = 0
cost = 5/3 не удовлетворяет условию: I cost I ≤ 1
8) !sin 3t! =1/2
a) sint = - 1/2
t = (-1)^(n)*arcsin( - 1/2) + πn, n∈Z
t = (-1)^(n+1)*arcsin(1/2) + πn, n∈Z
t1 = (-1)^(n+1)*(π/6) + πn, n∈Z
b) sint = 1/2
t = (-1)^(n)*arcsin(1/2) + πn, n∈Z
t2 = (-1)^(n)*(π/6) + πn, n∈Z