Solve x + x/sqrt(x^2 - 1) = 35/12 using trigonometric identities

[hearts123]

Hi everyone
Here's a question that I have to solve using trigonometry (identities). I have a vague idea of how to start, but not much else.

x + x/sqrt(x^2 - 1) = 35/12
the question wants me to substitute x = 1/cos α
α belongs to: (0, π/2)
I have to solve the equation, aka what are the possible values for x?

 

[MarkFL]

Yes, when I saw the thread title, I knew a substitution of the form:

[MATH]x=\sec(\alpha)[/MATH]
would be in order. So, we make the substitution:

[MATH]\sec(\alpha)+\frac{\sec(\alpha)}{\sqrt{\sec^2(\alpha)-1}}=\frac{35}{12}[/MATH]
Let's start with the radicand (expression under the radical)...can you think of an identity that will turn it into a square?

 

[hearts123]

Yes, when I saw the thread title, I knew a substitution of the form:

[MATH]x=\sec(\alpha)[/MATH]
would be in order. So, we make the substitution:

[MATH]\sec(\alpha)+\frac{\sec(\alpha)}{\sqrt{\sec^2(\alpha)-1}}=\frac{35}{12}[/MATH]
Let's start with the radicand (expression under the radical)...can you think of an identity that will turn it into a square?

tan^2(α) + 1 = sec^2(α)
So, I can turn sec^2(α) - 1 into tan^2(α) ?

 

[MarkFL]

tan^2(α) + 1 = sec^2(α)
So, I can turn sec^2(α) - 1 into tan^2(α) ?

Yes, exactly.

 

[hearts123]

Yes, exactly.

What I have now is sec(α) + sec(α) / tan(α) = 35/12
I can't think of any identities that I can use for that.
Could you give some hints for what I should do next?

 

[MarkFL]

I would simplify that to:

[MATH]\sec(\alpha)+\csc(\alpha)=\frac{35}{12}[/MATH]
Next, I think we should multiply by \(\sin(\alpha)\cos(\alpha)\) to get:

[MATH]\sin(\alpha)+\cos(\alpha)=\frac{35}{12}\sin(\alpha)\cos(\alpha)[/MATH]
Now, let's square both sides (we know both sides are positive...right?):

[MATH]\sin^2(\alpha)+2\sin(\alpha)\cos(\alpha)+\cos^2(\alpha)=\left(\frac{35}{12}\right)^2\sin^2(\alpha)\cos^2(\alpha)[/MATH]
Or:

[MATH]2\sin(\alpha)\cos(\alpha)+1=\left(\frac{35}{12}\right)^2\sin^2(\alpha)\cos^2(\alpha)[/MATH]
Or:

[MATH]\left(\frac{35}{24}\right)^2\sin^2(2\alpha)-\sin(2\alpha)-1=0[/MATH]
Now you have a quadratic in \(\sin(2\alpha)\)...

 

[MarkFL]

To follow up...

Solving the above quadratic, we find:

[MATH]\sin(2\alpha)\in\left\{-\frac{24}{49},\frac{24}{25}\right\}[/MATH]
Now, since \(0<2\alpha<\pi\) we only consider:

[MATH]\sin(2\alpha)=\frac{24}{25}[/MATH]
And so we find :

[MATH]x=\sec\left(\frac{1}{2}\arcsin\left(\frac{24}{25}\right)\right)=\frac{5}{4}[/MATH]
[MATH]x=\sec\left(\frac{1}{2}\left(\arcsin\left(\frac{24}{25}\right)-\pi\right)\right)=\frac{5}{3}[/MATH]
Thus, we conclude:

[MATH]x\in\left\{\frac{5}{4},\frac{5}{3}\right\}[/MATH]

 

[topsquark]

I can see that it can be done. But given how annoying the derivation is, why would you actually want to do it this way? ?

-Dan