4) Sin⁴x + Cos⁴x + Sin2x = 7/5
(Sin²x)² + Cos⁴x + Sin2x = 7/5
(1 - Cos²x)² + Cos⁴x + Sin2x = 7/5
1 - 2Cos²x + Cos⁴x +Cos⁴x + Sin2x = 7/5
-2Cos²x +2Cos⁴x + Sin2x = 2/5
-2Cos²x +2Cos⁴x +2SinxCosx = 2/5
-Cos²x +Cos⁴x +SinxCosx = 1/5
-Cos²x(1 - Cos²x) + SinxCosx = 1/5
-Cos²xSin²x + SinxCosx = 1/5 |*4
-4Cos²xSin²x + 2*2SinxCosx = 4/5
-Sin²2x +2Sin2x = 4/5 |*5
-5Sin²2x + 10Sin2x -4 = 0
Sin2x = t
-5t² + 10t - 4 = 0
5t² -10t +4 = 0
t = (5+-√(25 -20))/5
t₁ = (5+√5)/5= 1 + √5/5 t₂= 1 - √5/5
Sin2x = 1 + √5/5 Sin2x = 1 - √5/5
∅ 2x = (-1)^n arcSin (1 - √5/5) + nπ, n ∈Z
x = (-1)^n arcSin(1 - √5/5)/2+ nπ/2, n ∈Z
5) (х² -х +1)⁴ - 5х²(х² -х +1)² + 4х⁴ = 0
(х² - х +1)² = t
t² - 5x² t + 4x⁴ = 0
D = b² -4ac = 25x⁴ - 16x⁴ = 9x⁴
t₁= (5x² + 3x²)/2 = 4x² t₂ = (5x² - 3x²)/2 = х²
(х² - х +1)² = t
(х² - х +1)² = 4х² (х² - х +1)² = х²
а)х²-х +1 = 2х б) х² -х +1 = -2х в) х² -х +1 = х г) х² -х +1 = -х
х² -3х +1 = 0 х² +х +1 = 0 х²-2х +1 =0 х² +1 = 0
D = 5 D = -3 (x-1)² =0 x² = -1
x = (3 +-√5)/2 ∅ х = 1 ∅
Ответ: x = (3 +-√5)/2
х = 1
Х^2+2х-3=0
х^2-7х-5=0
2х^2+22х-17=0
х^2-4-3х=0 или х^2-3х-4=0
3(х^2-1)+х=0 ; 3х^2+х-3=0
х^2-2х+3х-6=0 ; х^2+х-6=0
(-0,6x²-2)/(1-2y)
x=-2 y=0,3
(-0,6*4-2)/(1-2*0,3)=(-2,4-2)/(1-0,6)=(-4,4)/0,4=-11
Вот там надо (1-2в)(1+2в+4в в квадрате)