Помогите с уравнениями пожалуйста!20б!! 1) 2^(sin^2x)

alinagunba99 @alinagunba99

July 2022 1 4 Report

Помогите с уравнениями пожалуйста!20б!!

1) 2^(sin^2x) - 5* 2^(cos^2x) = 7

2) 3^x - 5* 3^(-x) >=4

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sharadi

Verified answer

2^(Sin²x) + 5*2^(1 - Sin²x) = 7
2^Sin²x + 5*(2*2^-Sin²x) = 7
2^Sin²x + 10*2^-Sin²x = 7|*2^Sin²x
(2^Sin²x)² +10 -7*2^Sin²x = 0
2^Sin²x = t
t² - 7t +10 = 0
корни 2 и 5
2^Sin²x = 2 или 2^Sin²x = 5
Sin²x = 1   ∅
Sinx = +-1
x = +-π/2 + πk , k ∈Z

2) 3^x - 5* 3^(-x) ≥ 4 | * 3^x
(3^x)² -5 -4*3^x ≥ 0
3^x = t
t² -4t -5 ≥ 0 ( корни 5 и -1)
t ≤ -1   или t ≥ 5
3^x ≤ -1 3^x ≥ 5
∅ xlg3 ≥ lg5
x ≥ lg5/lg3

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Answer

alinagunba99 August 2022 | 0 Ответы

Answers & Comments

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