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Arinak98
@Arinak98
July 2022
2
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1.Решить уравнение
2sin x-√3=0
2. Вычислить используя формулы приведения
Cos330(градусов)
3. Найти производную функции y=(x+1)(x^3-x) при x=2
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Answers & Comments
nastyam9790
1) 2sinx = √3
sinx = √3/2
x = π/3 + 2πk
x = 2π/3 + 2πk
2) cos(330) = cos(360 - 30) = cos(2π - π/6) = cos(π/6) = √3/2
3) y = x^4 - x^2 + x^3 - x = x^4 + x^3 - x^2 - x
y' = 4x^3 + 3x^2 - 2x - 1
x=2, y'(2) = 4*8 + 3*4 - 4 - 1 = 39
4 votes
Thanks 11
nastyam9790
2.cos(330)=cos(2pi-30)=cos(30)
kalbim
Verified answer
1.Решение:
2 sinx - √3 =0
sinx = √3 /2
х = (-1)^n π/3 +πn, n∈Z .
(-1) в степени "n"
3 votes
Thanks 8
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Answers & Comments
sinx = √3/2
x = π/3 + 2πk
x = 2π/3 + 2πk
2) cos(330) = cos(360 - 30) = cos(2π - π/6) = cos(π/6) = √3/2
3) y = x^4 - x^2 + x^3 - x = x^4 + x^3 - x^2 - x
y' = 4x^3 + 3x^2 - 2x - 1
x=2, y'(2) = 4*8 + 3*4 - 4 - 1 = 39
Verified answer
1.Решение:2 sinx - √3 =0
sinx = √3 /2
х = (-1)^n π/3 +πn, n∈Z .
(-1) в степени "n"