Solve the equation 4-6coshx + 3-2sinhx = 10

[dannys9696]

Can anyone help me out with this equation...?

Solve the equation 4-6coshx + 3-2sinhx = 10
Giving the answer in 4 decimal place.

Just need help getting me on my way so i can understand thanks

 

[dannys9696]

Can anyone help me out with this equation...?

Solve the equation 4-6coshx + 3-2sinhx = 10
Giving the answer in 4 decimal place.

Just need help getting me on my way so i can understand thanks

 

[pka]

Solve the equation 4-6coshx + 3-2sinhx = 10

I cannot help but think that you posted the wrong form of this question.
As posted here it is:
\(\displaystyle 4-6\cosh(x)+3-2\sinh(x)=10\\6\cosh(x)+2\sinh(x)=-3\\6\dfrac{e^x+e^{-x}}{2}+2\dfrac{e^x-e^{-x}}{2}=-3\\3(e^x+e^{-x})+(e^x-e^{-x})=-3\)

 

[Ishuda]

Can anyone help me out with this equation...?

Solve the equation 4-6coshx + 3-2sinhx = 10
Giving the answer in 4 decimal place.

Just need help getting me on my way so i can understand thanks

If you don't want to go back to the basic definition [which I would also] and instead use the hyperbolic trig identities, you could do something like the following example [I'm not exactly sure of what you have]: Solve the equation
coshx + a sinhx = b.
Letting c=cosh(x) and s=sinh(x) this becomes
c = b - a s
Square both sides
c² = b² -2ab s + a² s²
Use the hyperbolic trig idenity
c² = 1 + s²
to get
...
Continue from there

Note that squaring to produce an answer sometimes introduces a spurious answer (especially if one is not careful) such proving not number but zero can be equal to itself. Proof is by contradiction: Suppose a and b are non-zero and equal. Then, since a=b, if we square each side,
a² = b²
or
a² - b² = (a-b) (a+b) = 0
Dividing by a-b we get (the spurious solution because I was not careful)
a+b = 0
or a=-b. Now, if a=b and a=-b then both a and b must both be zero which contradicts our premise that a and b are non-zero.

 

[dannys9696]

I cannot help but think that you posted the wrong form of this question.
As posted here it is:
\(\displaystyle 4-6\cosh(x)+3-2\sinh(x)=10\\6\cosh(x)+2\sinh(x)=-3\\6\dfrac{e^x+e^{-x}}{2}+2\dfrac{e^x-e^{-x}}{2}=-3\\3(e^x+e^{-x})+(e^x-e^{-x})=-3\)

i think the equation is supposed to be 4.6coshx + 3.2sinhx = 10
thanks for the help