Решение
3cos³x - 7cos²xsinx + 4sin³x = 0 делим на cos³x ≠ 0
4 tg³x - 7tgx + 3 = 0
tgx = t
4t³ - 7t + 3 = 0
t₁ = 1
4t³ - 7t + 3 It - 1I
-<u>(4t³ - 4t²) </u> 4t² + 4t - 3
4t² - 7t
-<u>(4t² - 4t)</u>
- 3t + 3
-<u>( -3t + 3)</u>
0
4t³ - 7t + 3 = (t - 1)*(4t² + 4t - 3)
4t² + 4t - 3 = 0
D = 16 + 4*4*3 = 64
t = (- 4 - 8)/8
t₂ = - 1,5
t = (- 4 + 8)/8
t₃ = 0,5
1) tgx = 1
x₁ = π/4 + πm, m∈Z
2) tgx = - 1,5
x₂ = - arctg1,5 + πk, k∈Z
3) tgx = 1/2
x₃ = arctg(1/2) + πn, n∈ Z
33+3(6х-5у)^2+3(x+2y)^2-5y= 36(36x^2-60xy+25y)+3(2x^2+4xy+4y)-5y =
1296x^2-2160xy+900y+6x^2+12xy+12y-5y=1302x^2-2148xy+907y
Округление будет равно 1.5 погрешность, на +- 0.01 асбалютная, относительная +-0.1
1) cos(α+β)+2sinαsinβ=cosαcosβ-sinαsinβ+2sinααsinβ=cosαcosβ-sinαsinβ=cos(α+β)
2) (1+tg²α)(1-cos²α)=1/cos²α × sin²α=cos²α/sin²α=tg²α