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
1)log27(3+log2(x+2))=0
log27(3+log2(x+2))=log27 1
3+log2(x+2)=1
3 log2 2+log2(x+2)=log2 2
log2(2^3)+log2(x+2)=log2 2
log2( 8(·x+2)=log2 2 ОДЗ : х+2>0 x>-2
8(х+2)=2
8х+16=2
8х=2-16
8х=-14
х=-14:8
х=-1,75 -1,75>-2 (ОДЗ)
Ответ:-1,75
2) log3² (x)-3log3(x)=-10^lg2
1\2log3(x)-log3(x³)=-2
log3(√x)\x³=-2log3 3 ОДЗ:х>0
√x\x³=1\9
9√x=-x³
-x²√x=9
x^(5|2)=-9 корней нет ( возможно что то в условии было непонятно)
3) log(x+2) (3x²-12)=2
log(x+2) (3x²-12)=log(x+2) (x+2) ОДЗ: х+2≠1 х≠-1 и х+2>0 x>-2
3x²-12=x+2
3x²-x-14=0
D=1-4·3·(-14)=1+168=169 √D=13
x1=(1+13)\6=7\3=2 1\3
x2=(1-13)\6=-12\6=-2 ( не является корнем , ОДЗ исключает )
Ответ: х=2 1\3
5)log2 (2x-3)+ log2 (1-x)=1
log2 (2x+3)(1-x)=log2 2 ОДЗ:2х+3>0 2x>-3 x>-1.5
1-x>0 -x>-1 x<1
2x+3)(1-x)=2
2x-2x²+3-3x-2=0
2x²+x-1=0
D=1-4·2·(-1)=9 √D=3
x1=(-1+3)\4=1\2
x2=(-1-3)\4=-1
x1·x2=-1·1\2=-1\2
6) log2 x+ logx 16=5 Одз: х≠1 х>0
log 2 x+ 1\(log16 x)=5
log2 x+1\(log2^4 (x))=5
log2 x +4\(log2 x)=5
log² 2 x+4 -5log2 x=0
введём замену переменной , пусть log2 x=y
y²-5y+4=0
D=25-4·4=9 √D=3
y1=(5+3)\2=4
y2=(5-3)\2=1
возвращаемся к замене:
log2 x=4
x=2^4=16
log2 x=1
x=2
x1+x2=16+2=18
условие примера 4 не совсем точно понимаю, уточните
Х(2х+72)=0
х=0 или 2х+72=0
х= - 72/2
х= -36
ответ 0, -36
