............................
Сos(x)+√((2-√2)/2*(sin(x)+1))=0
сos(x)=-√((2-√2)/2*(sin(x)+1))
√(1-sin²(x))=-√((1-√2/2)*(sin(x)+1))
1-sin²(x)=(1-√2/2)*(sin(x)+1)
1-sin²(x)=1-√2/2 + sin(x) - √2/2*sin(x)
sin²(x) + sin(x)-√2/2*sin(x) - √2/2=0
sin(x)*(sin(x)+1)-√2/2*(sin(x)+1)=0
(sin(x)-√2/2)*(sin(x)+1)=0
1. sin(x)-√2/2=0
sin(x)=√2/2
Проверка:
√2/2+√((2-√2)/2*(√2/2+1))=0
√2/2+√((1-√2/2)*(√2/2+1))=0
√2/2+1-√2/2=0
1≠0
Посторонний корень.
2. sin(x)+1=0
sin(x)=-1
Проверка:
0+√((2-√2)/2*(-1+1))=0
√0=0
Корень является решением данного уравнения
х=arcsin(-1)+ 2*π*n
x=(3π)/2+2πn
Ответ: x=(3π)/2+2πn
(10x + 45)e^(1-x) + (5x^2 + 45x - 45)e^(1-x)(-1) = e^(1-x) (10x + 45 - 5x^2 - 45x + 45) = e^(1-x)(-5x^2 - 35x + 90)
-5x^2-35x+90 = 0
X^2 + 7x - 18 = 0
D = 49 + 72 = 121
x = (-7 + 11)/2 = 2
x = (-7 - 11)/2 = -9
- + -
____-9_______2___>
Min. Max
Ответ: -9
Решение<span>
cos2x-cos^2x-2cosx<-2
2cos²x - 1 - cos²x - 2cosx < - 2
cos²x -
2cosx + 1 < 0
cosx = t
t² - 2t +
1 = 0
(t - 1)²
= 0
t = 1
cosx <
1
</span><span>2πn </span><span>< x < 2π + 2πn, n </span><span>∈ Z</span>