Объяснение:
( x + 2 ) ^ 4 - 4 * ( x + 2 ) ^ 2 - 5 = 0 ;
Пусть ( х + 2 ) ^ 2 = а, тогда:
а ^ 2 - 4 * a - 5 = 0 ;
a1 = ( 4 - √36 ) / ( 2 * 1 ) = ( 4 - 6 ) / 2 = - 2 / 2 = - 1 ;
a2 = ( 4 + √36 ) / ( 2 * 1 ) = ( 4 + 6 ) / 2 = 10 / 2 = 5 ;
Тогда:
1 ) ( x + 2 ) ^ 2 = - 1 ;
x ^ 2 + 4 * x + 4 = - 1 ;
x ^ 2 + 4 * x + 4 + 1 = 0 ;
x ^ 2 + 4 * x + 5 = 0 ;
Нет корней ;
2 ) ( x + 2 ) ^ 2 = 5 ;
x ^ 2 + 4 * x + 4 = 5 ;
x ^ 2 + 4 * x - 1 = 0 ;
x1 = ( -4 - √20 ) / ( 2·1 ) = -2 - √5 ;
x2 = ( -4 + √20 ) / ( 2·1 ) = -2 + √5 ;
ответ: х = -2 - √5 и х = -2 + √5
sin2x - (1-sin²x) =0 ;
2sinxcosx -cos²x =0 ;
cosx(2sinx -cosx) =0 ;
[cosx =0 ;2sinx-cosx =0.⇔ [cosx =0 ;sinx=(1/2)cosx.⇔[cosx =0 ;tqx=1/2.
[ x=π/2 +πn ; x =arctq1/2+πn , n∈Z.
2) ;
ctq2x*cos²x - ctq2x*sin²x =0 ;
ctq2x*(cos²x - sin²x) =0 ;
ctq2x*cos2x =0 ;
sin2x =0 * * *cos2x = ± 1 ≠0→ ОДЗ * * *
2x =πn , n∈Z ;
x =(π/2)*n , n∈Z .
3) ;
3sin²x/2 -2sinx/2 =0 ;
3sinx/2 (sinx/2 -2/3) =0 ;
[sinx/2 =0 ; sinx/2 =2/3 .⇒[x/2 =πn ; x/2= arcsin(2/3) +πn ,n∈Z.⇔
[x =2πn ; x= 2arcsin(2/3) +2πn ,n∈Z.
4) ;
* *cos2α =cos²α -sin²α =cos²α -(1-sin²α)=2cos²α -1⇒1+cos2α=2cos²α * *
cos3x = 1+cos2*(3x) ; * * * α = 3x * * *
cos3x = 2cos²3x ;
2cos²3x -cos3x =0 ;
2cos3x(cos3x -1/2) =0 ;
[cos3x =0 ; cos3x =1/2 ⇒[3x=π/2+πn ; 3x= ±π/3+2πn ,n∈Z.⇔
[x=π/6+πn/3 ; x= ±π/9+(2π/3)*n ,n∈Z.
Объяснение:
( x + 2 ) ^ 4 - 4 * ( x + 2 ) ^ 2 - 5 = 0 ;
Пусть ( х + 2 ) ^ 2 = а, тогда:
а ^ 2 - 4 * a - 5 = 0 ;
a1 = ( 4 - √36 ) / ( 2 * 1 ) = ( 4 - 6 ) / 2 = - 2 / 2 = - 1 ;
a2 = ( 4 + √36 ) / ( 2 * 1 ) = ( 4 + 6 ) / 2 = 10 / 2 = 5 ;
Тогда:
1 ) ( x + 2 ) ^ 2 = - 1 ;
x ^ 2 + 4 * x + 4 = - 1 ;
x ^ 2 + 4 * x + 4 + 1 = 0 ;
x ^ 2 + 4 * x + 5 = 0 ;
Нет корней ;
2 ) ( x + 2 ) ^ 2 = 5 ;
x ^ 2 + 4 * x + 4 = 5 ;
x ^ 2 + 4 * x - 1 = 0 ;
x1 = ( -4 - √20 ) / ( 2·1 ) = -2 - √5 ;
x2 = ( -4 + √20 ) / ( 2·1 ) = -2 + √5 ;
ответ: х = -2 - √5 и х = -2 + √5