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Sorvigolova57
@Sorvigolova57
July 2022
1
15
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Решите логарифмическое неравенство:
Lg(2x+1)<0
Решите тригонометрическое уравнение:
2*соs(x/4)-корень3=0
Найти производную:
f(x)=sin^2x
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sharadi
Verified answer
1) lg(2x +1) < lg1
2x +1 > 0 x > -1/2
2x +1 <1, ⇒ x < 0
Ответ: х∈(-1/2; 0)
2) 2Cos(x/4) = √3
Cos(x/4) = √3/2
x/4 = +-arcCos√3/2 + 2πk , k ∈Z
x/4 = +-π/6 + 2πk , k ∈ Z
x = +-2π/3 + 8πk , k ∈Z
3) f(x) = Sin²x
f'(x) = 2Sinx * (Sinx)' = 2SinxCosx = Sin2x
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Answers & Comments
Verified answer
1) lg(2x +1) < lg12x +1 > 0 x > -1/2
2x +1 <1, ⇒ x < 0
Ответ: х∈(-1/2; 0)
2) 2Cos(x/4) = √3
Cos(x/4) = √3/2
x/4 = +-arcCos√3/2 + 2πk , k ∈Z
x/4 = +-π/6 + 2πk , k ∈ Z
x = +-2π/3 + 8πk , k ∈Z
3) f(x) = Sin²x
f'(x) = 2Sinx * (Sinx)' = 2SinxCosx = Sin2x