4sin^2(x)+4cos(x)-5=0
По формуле sin^2(x)=1-cos^2(x):
4(1-cos^2(x))+4cos(x)-5=0
4-4cos^2(x)+4cos(x)-5=0
-4(cos(x))^2+4cos(x)-1=0
Сделаем замену переменной cos(x)=t:
-4t^2+4t-1=0 | *(-1)
4t^2-4t+1=0
D=b^2-4ac=(-4)^2-4*4*1=16-16=0
t=-b/2a=4/8=1/2
Сделаем обратную замену:
cos(x)=1/2
cos(α) = cos(2π - α) ⇒ cos(x) = 1/2 или cos(2π - x) = 1/2
1) x = arccos(1/2)
*** arccos(1/2) = π/3 ***
x = π/3
x = π/3 + 2πn, n ∈ Z
2) 2π - x = arccos(1/2)
2π - x = π/3
- x = π/3 - 2π
- x = (π - 6π)/3
- x = - 5π/3
- x = - 5π/3 + 2πn, n ∈ Z
x = 5π/3 - 2πn, n ∈ Z
ответ: x = π/3 + 2πn, n ∈ Z
x = 5π/3 - 2πn, n ∈ Z
2х-х>6+8
х>14