Question

How do you solve `7 sinh x - 5 cosh x=7`?

Answer 1

90^@ and 161^@08

Explanation:

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

Answer 2

`x = ln ((7+sqrt73)/2)= 2.0505`, nearly.

Explanation:

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)`.