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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

Jack D'Aurizio
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mja
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    $p(t)=8t^3-4t^2-8t+3$ has three real roots, but just two of them lie in $[-1,1]$. You may compute the exact roots through Cardano's formula. – Jack D'Aurizio Aug 20 '15 at 07:56
  • Thanks. I know the cubic equation has 3 real roots and I had these approximate roots, but I think it is very complex to find these roots when my school does not allow me using the Cardano's formula. – mja Aug 20 '15 at 08:10

2 Answers2

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HINT:

Notice, we have $$\cos \frac{4x}{3}=\cos x+1$$ Let, $x=3t$ , we get $$\cos 4t=\cos 3t+1$$ $$2\cos^2 2t-1=4\cos^3 t-3\cos t+1$$ $$2(2\cos^2 t-1)^2-4\cos^3 t+3\cos t-2=0$$ $$8\cos^4 t-8\cos^2t+2-4\cos^3 t+3\cos t-2=0$$ $$\cos t(8\cos^3 t-4\cos^2t+2-8\cos t+3)=0$$

$$\iff \cos t=0\iff t=\frac{\pi}{2}\iff x=\frac{3\pi}{2}$$

$$\iff 8\cos^3 t-4\cos^2t+2-8\cos t+3=0$$ Can you solve this cubic equation for $\cos t$?

Harish Chandra Rajpoot
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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.

Community
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Claude Leibovici
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