If $\cos \alpha\;,\cos \beta\;,\cos \gamma$ are the roots of $9x^3-9x^2-x+1=0\;, $ Then $\alpha+\beta+\gamma$

Question

If $\cos \alpha\;,\cos \beta\;,\cos \gamma$ are the roots of the equation $9x^3-9x^2-x+1=0\;, \alpha,\beta,\gamma \in \left[0,\pi\right]$

Then value of $ \left(\sum \alpha\;,\sum \cos \alpha\right)$

$\bf{My\; Try::}$ Using Vieta formula::

$$\cos \alpha+\cos \beta+\cos \gamma = 1$$ and $$\cos \alpha\cdot \cos \beta+\cos \beta\cdot \cos \gamma+\cos \gamma\cdot \cos \alpha = -\frac{1}{9}$$

and $$\cos \alpha\cdot \cos \beta\cdot \cos \gamma = -\frac{1}{9}$$

Now How can i calculate $\alpha+\beta+\gamma = $, Help required, Thanks

Answer 1

HINT:

Clearly, $$0=9x^2(x-1)-(x-1)=(x-1)(9x^2-1)$$

Use How do I prove that $\arccos(x) + \arccos(-x)=\pi$ when $x \in [-1,1]$?

Answer 2

$x=1$ is solution, you can solve the equation