How do you solve #7 sinh x - 5 cosh x=7#?
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7sin x - 5cos x = 7
Divide both side by 7
#sin x - (5/7)cos x = 1#
Call #tan a = (5/7)# --> arc #a = 35^@54#
#sin x - (sin a)/(cos a)cos x = 1#
#sin x.cos a - sin a.cos x = cos a = cos 35.54 = 0.81#
#sin (x - a) = sin (x - 35.54) = 0.81#
There are 2 solutions:
a. #x - 35.54 = 54.46# --> #x = 54.46 + 35.54 = 90^@#
b. #x - 35.54 = 180 - 54.46 = 125.54# -->
#x = 125.54 + 35.54 = 161^@08#
Check by calculator.
#x = 90^@ #--> 7sin x = 7 --> -5cos x = 0 --> 7 = 7 .OK
#x = 161^@08# --> 7sin x = 2.28 --> -5cos x = 4.72 -->
2.28 + 4.72 = 7. OK
# x = ln ((7+sqrt73)/2)= 2.0505#, nearly.
Use #sinh x=(e^x-e^(-x))/2 and cosh x=(e^x+e^(-x))/2#.
Accordingly, #7((e^x-e^(-x))/2)-5((e^x+e^(-x))/2)=7#.
Multiply by #e^x#. The result is the quadratic in #e^x#,
[Thanks to Olthe3rd for duly pointing out the swapping of 6 an 7 below, in my previous edition, that took me to the wrong answer]..
#e^(2x)-7e^x-6=0#..
The roots are #e^x =(7+-sqrt73)/2.#.
As #e^x>=0#, the negative root is inadmissible.
So, #e^x=(7+sqrt73)/2#. Inverting, #x = ln( (7+sqrt73)/2)#.
