Решение
(sinx + 1)/(1 - cos2x) = (sinx + 1)/(1 + cos(π/2 + x))
<span>(sinx + 1)/(1 - cos2x) = (sinx + 1)/(1 + sinx)
</span><span>(sinx + 1)/(1 - cos2x) = 1
</span>
sinx + 1 = 1 - cos2x
1 - cos2x ≠ 0, cos2x ≠ 1, 2x ≠ 2πk, k ∈Z; x ≠ <span>πk, k ∈Z
</span>
sinx + cos2x = 0
sinx + 1 - 2sin²x = 0
2sin²x - sinx - 1 = 0
sinx = t
2t² - t - 1 = 0
D = 1 + 4*2*1 = 9
t₁ = (1 - 3)/4
t₁ = - 1/2
t₂ (1 + 3)/4
t₂ = 1
1) sinx = - 1/2
x = (-1)^n*arcsin(-1/2) + πn, n ∈ Z
x₁ = <span>(-1)^n* arcsin(-1/2) + πn, n ∈ Z
</span>x₁ = <span>(-1)^(n+1)* arcsin(1/2) + πn, n ∈ Z
</span>x₁ = (-1)^(n+1)* (π/6)<span> + πn, n ∈ Z
</span>2) sinx = 1
<span>x₂ = </span> π/2 + 2πm, m ∈ Z
<span>(9x^6 - 4x^3) - (x^3 - 9) - (8x^6 - 5x^3) = x^6 +9
</span><span>9x^6 - 4x^3 - x^3 +9 - 8x^6 +5x^3 = x^6 +9
находим подобные слагаемые (</span>9x^6 -8x^6 =x^6 ..........- 4x^3 - x^3 +<span>5x^3=0)
</span>
x^6 +9=<span>x^6 +9</span>
А)y^5-25y³=y^3(y-5)(y+5)
<span>б)16x+8x²+x³=x(x+4)^2</span>
-1/4=0,25;
-0,25*(-0,8)+0,16/(-0,8)=0,2+(-0,2)=0