Sin4x - cos⁴x = -sin⁴x
sin4x = cos⁴x - sin⁴x
sin4x = (cos²x - sin²x)(cos²x + sin²x)
sin4x = cos²x - sin²x
sin4x = cos2x
2sin2xcos2x - cos2x = 0
cos2x(2sin2x - 1) = 0
1) cos2x = 0
2x = π/2 + πn, n ∈ Z
x = π/4 + πn/2, n ∈ Z
2) 2sin2x - 1 = 0
sin2x = 1/2
2x = (-1)ⁿπ/6 + πk, k ∈ Z
x = (-1)ⁿπ/12 + πk/2, k ∈ Z
Ответ: x = π/4 + πn/2, n ∈ Z; (-1)ⁿπ/12 + πk/2, k ∈ Z.
D = a2 -a1 = 4-1 = 3
A(n) = a1 + (D*n-1)
A8 = a1 + ( D*(8-1)) = a1 + 21 = 22
A1+...A8 = 1+4+7+10 +13+16+19+22= 92
Х = у + 4
<span>Подставляем во второе уравнение </span>
<span>24 * (1/(у+4) + 1/у) = 5 </span>
<span>24 * (2у + 4) / (y^2 + 4y) = 5 </span>
<span>48y + 96 = 5y^2 + 20y </span>
<span>5y^2 - 28y - 96 = 0 </span>
<span>y1 = 8; х1 = 12 </span>
<span>у2 = -2,4; х2 = 1,6</span>