\(\cos\left(\frac{\pi}{6}\left(4x+\sqrt{10+x^2}\right)\right)=-\frac{\sqrt{3}}{2}\)
\(\cos\left(\frac{\pi}{6}\left(4x+\sqrt{10+x^2}\right)\right)=\cos\left(\frac{5\pi}{6}\right)\)
\(\frac{\pi}{6}\left(4x+\sqrt{10+x^2}\right)=\frac{5\pi}{6}\)
\(4x+\sqrt{10+x^2}=5\)
\(\sqrt{10+x^2}=5-4x\)
\(10+x^2=25-40x+16x^2\)
\(15-40x+15x^2=0\)
\(\sqrt{\Delta}=10\sqrt{7}\)
\(\orbr{\begin{cases}x_1=\frac{40+10\sqrt{7}}{30}=\frac{4+\sqrt{7}}{3}\left(ktm\right)\\x_2=\frac{40-10\sqrt{7}}{30}=\frac{4-\sqrt{7}}{3}\left(tm\right)\end{cases}}\)
vậy pt có n0 duy nhất là \(\frac{4-\sqrt{7}}{3}\)