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nadirabaratova
@nadirabaratova
August 2022
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3
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Помогите решить 2 задания, очень срочное :
1 задание:
Решите уравнение
sin x+sin 2x=tg x
2 задание:
Найдите производную
f(x)=(x-корень 1-Х)-корень х/3
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nKrynka
Решение
1 задание:
Решите уравнение
sin x+sin 2x=tg x
sinx + 2sinxcosx - sinx/cosx = 0 умножим на cosx
≠0, x ≠ π/2 + πk, k ∈ Z
sinxcosx + 2sinxcos²x - sinx = 0
sinx(cosx + 2cos²x - 1) = 0
1.
sinx = 0
x
₁
=
πn, n ∈ Z
2.
2cos
²x + cosx - 1 = 0
cosx = t
2t
² + t - 1 = 0
D = 1 + 4*2*1 = 9
t
₁
= (- 1 - 3)/4 = -1
t
₂
= (- 1 + 3)/4 = 1/2
1) cosx = - 1
x
₂
=
π + 2πk, k ∈ Z
2) cosx = 1/2
x = +-arccos(1/2) + 2
πm, m ∈ Z
x₂ = +-(π/3) + 2
πm, m ∈ Z
2 задание:
Найдите производную
f(x)=(x -
√(
1 - x) -
√
х/3
f`(x) = 1 + 1/[2
√(1 - x)] - 1 /[ 6√x]
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Answers & Comments
1 задание:
Решите уравнение
sin x+sin 2x=tg x
sinx + 2sinxcosx - sinx/cosx = 0 умножим на cosx≠0, x ≠ π/2 + πk, k ∈ Z
sinxcosx + 2sinxcos²x - sinx = 0
sinx(cosx + 2cos²x - 1) = 0
1.
sinx = 0
x₁ = πn, n ∈ Z
2.
2cos²x + cosx - 1 = 0
cosx = t
2t² + t - 1 = 0
D = 1 + 4*2*1 = 9
t₁ = (- 1 - 3)/4 = -1
t₂ = (- 1 + 3)/4 = 1/2
1) cosx = - 1
x₂ = π + 2πk, k ∈ Z
2) cosx = 1/2
x = +-arccos(1/2) + 2πm, m ∈ Z
x₂ = +-(π/3) + 2πm, m ∈ Z
2 задание:
Найдите производную
f(x)=(x - √(1 - x) - √х/3
f`(x) = 1 + 1/[2√(1 - x)] - 1 /[ 6√x]