giải pt $\sqrt[3]{6cosx + 2}$ = 2cos3x + 2cosx -2

Giải thích các bước giải:

$\sqrt[3]{6\cos x+2}=2\cos 3x+2\cos x-2$ 

$\to \sqrt[3]{6\cos x+2}=2(4\cos^3x-3\cos x)+2\cos x-2$ 

$\to \sqrt[3]{6\cos x+2}=8\cos^3x-4\cos x-2$ 

$\to (6\cos x+2)+\sqrt[3]{6\cos x+2}=(2\cos x)^3+2\cos x$ 

$\to (6\cos x+2)-(2\cos x)^3+\sqrt[3]{6\cos x+2}-2\cos x=0$ 

$\to (\sqrt[3]{6\cos x+2}-2\cos x)(\sqrt[3]{6\cos x+2}^2+\sqrt[3]{6\cos x+2}.2\cos x+4\cos^2x)+\sqrt[3]{6\cos x+2}-2\cos x=0$ 

$\to (\sqrt[3]{6\cos x+2}-2\cos x)(\sqrt[3]{6\cos x+2}^2+\sqrt[3]{6\cos x+2}.2\cos x+4\cos^2x+1)=0$ 

$\to \sqrt[3]{6\cos x+2}-2\cos x=0$

$\to \sqrt[3]{6\cos x+2}=2\cos x$

$\to 6\cos x+2=8\cos^3x$

$\to 8\cos^3x-6\cos x-2=0$

$\to 2(\cos x-1)(2\cos x+1)^2=0$

$\to \cos x=1\to x=2k\pi$

Hoặc $\cos x=\dfrac{-1}{2}$

$\to x=\dfrac{2\pi}{3}+k2\pi$ hoặc $x=\dfrac{4\pi}{3}+k2\pi$