Asked by Alice
solve the equation 5coshx + 3sinhx = 4
Answers
Answered by
Reiny
sinhx = (e^x - e^-x)/2
and
coshx = (e^x + e^-x)/ , by definition.
then:
5coshx + 3sinhx = 4
5(e^x + e^-x)/2 + 3(e^x - e^-x)/2 = 4
5e^x + 5e^-x + 3e^x - 3e^-x = 8
8e^x + 2e^-x = 8
times e^x
8 (e^x)^2 + 2 = 8e^x
a quadratic, let a = e^x
8a^2 - 8a + 2 = 0
4a^2 - 4a + 1 = 0
(2a - 1)^2 = 0
a = 1/2
then e^x = 1/2
x = ln(1/2) = ln1 - ln2= -ln 2
x = -ln 2
and
coshx = (e^x + e^-x)/ , by definition.
then:
5coshx + 3sinhx = 4
5(e^x + e^-x)/2 + 3(e^x - e^-x)/2 = 4
5e^x + 5e^-x + 3e^x - 3e^-x = 8
8e^x + 2e^-x = 8
times e^x
8 (e^x)^2 + 2 = 8e^x
a quadratic, let a = e^x
8a^2 - 8a + 2 = 0
4a^2 - 4a + 1 = 0
(2a - 1)^2 = 0
a = 1/2
then e^x = 1/2
x = ln(1/2) = ln1 - ln2= -ln 2
x = -ln 2
Answered by
Naty
Yeah amazing
Answered by
Mohammed
5coshx+3sinhx=4
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