Solve the equation 2coshx + sinhx = 2 given that x is a real number?
回答 (3)
2016-08-02 4:01 pm
2coshx + sinhx = 2
2[(eˣ + e⁻ˣ)/2] + [(eˣ - e⁻ˣ)/2] = 2
2(eˣ + e⁻ˣ) + (eˣ - e⁻ˣ) = 4
2eˣ + 2e⁻ˣ + eˣ - e⁻ˣ = 4
3eˣ + e⁻ˣ = 4
[3eˣ + (1/eˣ)]eˣ = 4eˣ
3(eˣ)² + 1 = 4eˣ
3(eˣ)² - 4eˣ + 1 = 0
(3eˣ - 1)(eˣ - 1) = 0
eˣ = 1/3 or eˣ = 1
ln(eˣ) = ln(1/3) or eˣ = e⁰
xln(e) = ln(1/3) or x = 0
x = ln(1/3) or x = 0
2[(eˣ + e⁻ˣ)/2] + [(eˣ - e⁻ˣ)/2] = 2
2(eˣ + e⁻ˣ) + (eˣ - e⁻ˣ) = 4
2eˣ + 2e⁻ˣ + eˣ - e⁻ˣ = 4
3eˣ + e⁻ˣ = 4
[3eˣ + (1/eˣ)]eˣ = 4eˣ
3(eˣ)² + 1 = 4eˣ
3(eˣ)² - 4eˣ + 1 = 0
(3eˣ - 1)(eˣ - 1) = 0
eˣ = 1/3 or eˣ = 1
ln(eˣ) = ln(1/3) or eˣ = e⁰
xln(e) = ln(1/3) or x = 0
x = ln(1/3) or x = 0
2016-08-02 4:09 pm
coshx => (eˣ + e⁻ˣ)/2 and sinhx => (eˣ - e⁻ˣ)/2
so, 2coshx + sinhx => eˣ + e⁻ˣ + eˣ/2 - e⁻ˣ/2 = 3eˣ/2 + e⁻ˣ/2
i.e. 3eˣ/2 + e⁻ˣ/2 = 2
=> 3eˣ + e⁻ˣ = 4
Then, multiplying by eˣ we get:
3e²ˣ + 1 = 4eˣ
i.e. 3e²ˣ - 4eˣ + 1 = 0
=> (3eˣ - 1)(eˣ - 1) = 0
so, eˣ = 1/3 or 1
i.e. x = ln(1/3) or 0
:)>
so, 2coshx + sinhx => eˣ + e⁻ˣ + eˣ/2 - e⁻ˣ/2 = 3eˣ/2 + e⁻ˣ/2
i.e. 3eˣ/2 + e⁻ˣ/2 = 2
=> 3eˣ + e⁻ˣ = 4
Then, multiplying by eˣ we get:
3e²ˣ + 1 = 4eˣ
i.e. 3e²ˣ - 4eˣ + 1 = 0
=> (3eˣ - 1)(eˣ - 1) = 0
so, eˣ = 1/3 or 1
i.e. x = ln(1/3) or 0
:)>
2016-08-02 3:52 pm
2 ( e^x+e^(-x) /2) + (e^x - e^(-x)) /2 = 2
e^x + 1/e^x + e^x /2 -1/ (2e^x) = 2
Multiply both sides by 2e^x
2e^(2x) + 2 + e^(2x) - 1 = 4e^x
3e^(2x) - 4 e^x + 1 = 0
Let y=e^x
3y^2-4y+1 = 0
3y^2-3y-y+1 =0
3y(y-1) -1(y-1) = 0
(y-1) (3y-1) = 0
y=1
e^x = 1
x = 0
3y=1
y = 1/3
e^x = 1/3
x = ln(1/3) = -ln(3)
x = 0, -ln(3)
e^x + 1/e^x + e^x /2 -1/ (2e^x) = 2
Multiply both sides by 2e^x
2e^(2x) + 2 + e^(2x) - 1 = 4e^x
3e^(2x) - 4 e^x + 1 = 0
Let y=e^x
3y^2-4y+1 = 0
3y^2-3y-y+1 =0
3y(y-1) -1(y-1) = 0
(y-1) (3y-1) = 0
y=1
e^x = 1
x = 0
3y=1
y = 1/3
e^x = 1/3
x = ln(1/3) = -ln(3)
x = 0, -ln(3)
收錄日期: 2021-04-18 15:22:58
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20160802074423AASjdrK