Solving: cos(k+x) + cos(k-x) = c

[bobie]

Homework Statement

Hi, (this is not homework)

is it possible to solve an equation like this:

cos( 30° + x) +3 cos (30° - x) [0] = √13 ?

I already know x = 16°.1
do you know how to find it?
Thanks

 

[mfb]

I guess the 0 after the second cos is a typo?
It is possible, you can simplify the equation to A*cos(x) = √13 with the right A, afterwards you can solve for all x (there is more than one solution).

 

[bobie]

I guess the 0 after the second cos is a typo?
It is possible, you can simplify the equation to A*cos(x) = √13 with the right A, afterwards you can solve for all x (there is more than one solution).

Yes it is a typo (= 0)

Thanks,
how do I choose the right A = 3.7527368 ?, and how do I choose the right solution among the many solutions?

 

[Mentallic]

Can you expand cos(A+B) and cos(A-B)?

 

[bobie]

cos30cosx -sin30*sinx + 3*(cos30cosx+sin30sinx) = √13 ?

 

[mfb]

That 3 is new... then you'll need an additional phase in the cosine (or use both sine and cosine).
Anyway, you can use the trigonometric identities to simplify the problem. And you can use a computer to check your work.

 

[bobie]

I thought there might be a simple way, just out of interest.
I do not want to use a computer. Probably it is a complex procedure
Thanks for your help

 

[Mentallic]

cos30cosx -sin30*sinx + 3*(cos30cosx+sin30sinx) = √13 ?

Where did that 3 in bold come from?

I thought there might be a simple way, just out of interest.
I do not want to use a computer. Probably it is a complex procedure
Thanks for your help

The issue with your problem is that $\sqrt{13}>2$ and $\cos{x}\leq 1$ for all real x values, so a sum of two cosines couldn't possibly be more than 2. Your answer would be imaginary.

 

[D H]

The issue with your problem is that $\sqrt{13}>2$ and $\cos{x}\leq 1$ for all real x values, so a sum of two cosines couldn't possibly be more than 2. Your answer would be imaginary.

Look at the original post. It now reads (emphasis mine) $\cos(30^{\circ}+x)+\mathbf{3}\cos(30^{\circ}-x) = \surd 13$, instead of the original $\cos(30^{\circ}+x)+\cos(30^{\circ}-x) = \surd 13$.

That factor of three makes a huge difference. The original problem did not have a solution in the reals. Add that factor of three in and real solutions do exist.bobie, when you make a mistake in formulating the question, common courtesy demands that you tell us about the mistake. There's no problem making a mistake and letting us know you made a mistake, at least if if you do so early enough. We all make mistakes, after all. There's a big problem with silently changing the formulation and not telling us about the change.

 

[D H]

cos30cosx -sin30*sinx + 3*(cos30cosx+sin30sinx) = √13 ?

Correct. The left-hand side can be simplified in two ways. One is to collect terms, the other is to replace cos(30°) and sin(30°) with their numeric values. What does that leave you with?

 

[bobie]

I am awfully sorry I dropped and corrected 3 in the OP .

I found
3.4641 cosx + sinx =√13

 

[D H]

That's approximately correct. You'll be much better off if you express that symbolically. Use $\frac{\surd 3}2$ instead of 0.8660254 as the cosine of 30°. This will make it possible to come up with a nice expression for your x.

 

[bobie]

2√3 cosx + sinx = √13, √12 cosx + sinx =√13

cos x = √13/12 - sinx /√12
sinx = √13 - √12 cosx

then
sinx =√1-cosx², or is there a better way?

 

[D H]

Yes, there's a better way. Start with 2√3 cosx + sinx = √13. Divide both sides by cos(x) and then square both sides. On the right hand side you'll have 13/cos²(x). What's another way to write this? (What's 1/cos(x)?) What trigonometric identity can you apply here?

Hint: Eventually you should get a very simple expression for the tangent of x.

 

[bobie]

tanx + √12= √13 secx

tanx² + 2√12 tanx= 13 secx² -12

2√12 tanx = 13 secx²- tanx² - 12
tanx = (13 secx²- tanx² - 12) /√48

 

[D H]

There's a key trig identity that relates secant and tangent.