<s>[INST] Solve $\cos \frac{4x}{3}=\cos x+1$ 
Solve the equation \begin{equation} \cos \frac{4x}{3}=\cos x+1\tag 1\end{equation}

I had tried by taking $\cos\dfrac x3=t$ and from this we have $\displaystyle\cos\frac{4x}3=2\left(2t^2-1\right)^2-1; \cos x=4t^3-3t$
$(1) \iff t\left(8t^3-4t^2-8t+3\right)=0$
But I can't solve $8t^3-4t^2-8t+3=0$ because it gives me the approximate roots when I need exact roots
 [/INST] There is a trigonometric method for solving cubic equations. Not very pleasant but doable. Applying the method to the case of $$8t^3-4t^2-8t+3=0$$ Using $A=-\frac{1}{2}$, $B=-1$, $C=\frac{3}{8}$, $Q=-\frac{13}{36}$, $R= -\frac{43}{432}$, $D=-\frac{257}{6912}$, $$\theta= \cos ^{-1}\left(-\frac{43}{26 \sqrt{13}}\right)$$ the solutions are then $$t_1=\frac{1}{6}+\frac{\sqrt{13}}{3}  \cos \left(\frac{\theta }{3}\right)$$ $$t_2=\frac{1}{6}+\frac{\sqrt{13}}{3}  \cos \left(\frac{\theta +2\pi}{3}\right)$$ $$t_3=\frac{1}{6}+\frac{\sqrt{13}}{3}  \cos \left(\frac{\theta +4\pi}{3}\right)$$As Jack D'Aurizio pointed it out, $t_1 \gt 1$ must be discarded.
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