Trig to Surd

hint: $\cos 3\theta = 4 \cos^3 \theta - 3 \cos \theta$

$\begin{aligned} \text{let }x=\cos(\frac{1}{3}\arccos \frac{11}{16}) \\ 4x^3-3x=\cos(\arccos \frac{11}{16})=\frac{11}{16} \\ 64x^3-48 x-11=0 \\ (4x+3)(16 x^2-4x-11)=0 \\ x=\frac{-1}{4} \text{ or } \frac{1-3\sqrt{5}}{8} \text{ or } \frac{1+3\sqrt{5}}{8} \\ \\ 0<\frac{11}{16}<1 \\ 0<\arccos \frac{11}{16}<\frac{\pi}{2} \\ 0<\frac{1}{3} \arccos \frac{11}{16}<\frac{\pi}{6} \\ \frac{1}{2}<\cos(\frac{1}{3} \arccos \frac{11}{16})<1 \\ \frac{1}{2}<x<1 \\ \therefore x = \frac{1+3\sqrt{5}}{8} \\ \therefore \frac{1-3\sqrt{5}}{8} = -0.713525 \end{aligned}$