Solutions of Polynomial equations (1 Viewer)

[ConfusionIO]

Hey guys, I've seen questions like these questions been asked a few times here but I still don't understand a few things about these questions.
The most general questions I've come across so far are questions like these: Solve 8x³-6x-1=0 using cos3θ = 4cos³θ - 3cosθ. Hence deduce cos(π/9) = cos(2π/9) + cos(4π/9). *Cambridge 4.4 Example 20

This is what I know so far:
We want to make the equation for x equal to the one in terms of cos: so 8x³-6x-1=0 and 4cos³θ - 3cosθ. The typical method would be to let x = cosθ. Next, by factorizing 8x³-6x-1=0 we can get 2(4x³-3x) -1=0 thus 2(4cos³θ-3cosθ) -1 = 0. So now we can use the cos3θ result to get 2(cos3θ) - 1 = 0. Arranging to get 3θ = 1/2. The next part is the one which confuses me, if I recall correctly it is the general solution to trigonometric equations from 3U Trig. I've learn doing this way seems to work to get me θ;

cos^-1 (1/2 ± 2πk)/3 which simplifies down to (π ± 6πk)/9. Now what I do is, since the original equation is of degree 3, we need values for k up to 3. So I get the following values for x = cosθ where θ = (π ± 6πk)/9 for values of k = 0, ± 1, ± 2, ± 3. And that way I get x = cos(π/9), cos(7π/9), cos(-5π/9), cos(13π/9), cos(-11π/9), cos(19π/9), cos(-17π/9). However the Cambridge has different x values to what I have. Furthermore they say that values of θ give 3 distinct values of cosθ which some like the cos(π/9) I got right but they make cos(5π/9) = -cos(4π/9), which I think they have subtracted from 180 (π), but why do this, what difference would it make? Any ways they do the same for cos(7π/9) making it -cos(4π/9) and then they list the three.

That is what I think the hardest part for me is to understand. After that, I have am clueless on how to deduce that cos(π/9) = cos(2π/9) + cos(4π/9).

Any help is greatly appreciated, I have 4U Exam coming up in a week! :I

 

[Carrotsticks]

You should only be substituting in 3 values of K, you are substituting in 7 values. The reason why is because each value of K corresponds to 1 solution for X. If you keep going, you're just cycling through them again.

The way you are expressing your answers is confusing (using inverse cos etc), try something a bit more like this.



It's not 'totally necessary' to do this, but the reason why the Q changes from cos(7pi/9) to - cos(2pi/9) is so then you can have a fairly neat result in the end which shows the relationships between the cosine of various angles.

All I did when I was solving the trig equation was start with the 'basic' solutions pi/3 and keep generating them (next one is 5pi/3) etc until I have 3 values.

You might be wondering why I didn't do the usual plus/minus stuff (ie: why didn't I take cos(plus/minus pi/3) as my first 2 solutions?)

The reason why is because I'm looking for 3 DISTINCT roots. If I were to do the plus/minus stuff, I wouldn't get distinct roots, since cos(X) = cos(-X).

 

[ConfusionIO]

How are the sum of the roots equal to 0? I think I'm missing something; what I'm thinking is sum of roots = -b/a.

 

[Carrotsticks]

The sum of roots are equal to zero because the polynomial is 8x^3 + 0x^2 -6x - 1 = 0.

Notice that the coefficient of x^2 is zero, hence the sum of roots (using -b/a) is -0/8 = 0.

 

[iStudent]

Thanks carrot!
I always wondered how they removed the plus or minus sign when solving cos* = 0 lol