Answer: [tex]\bold{d)\ \dfrac{5}{8},\ \dfrac{35}{8},\ \dfrac{45}{8},\ \dfrac{75}{8}}[/tex]
Step-by-step explanation:
[tex]20\cos\bigg(\dfrac{2\pi}{5}x\bigg)=10\sqrt2\\\\\\\cos\bigg(\dfrac{2\pi}{5}x\bigg)=\dfrac{\sqrt2}{2}\\\\\\\cos^{-1}\bigg[\cos\bigg(\dfrac{2\pi}{5}x\bigg)\bigg]=\cos^{-1}\bigg(\dfrac{\sqrt2}{2}\bigg)\\\\\\ \dfrac{2\pi}{5}x=\cos^{-1}\dfrac{\sqrt2}{2}\\\\\\\text{Where on the Unit Circle is cos}=\dfrac{\sqrt2}{2}\quad \rightarrow\quad at\ \dfrac{\pi}{4}\ and\ \dfrac{7\pi}{4}\\\\\\\text{All possible solutions are:}\ \dfrac{\pi}{4}+2\pi n\quad and\quad \dfrac{7\pi}{4}+2\pi n\\\\\text{:n = number of rotations}[/tex]
[tex]\dfrac{2\pi}{5}x=\dfrac{\pi}{4}\quad \longrightarrow \quad \bigg(\dfrac{5}{2\pi}\bigg)\dfrac{2\pi}{5}x=\dfrac{\pi}{4}\bigg(\dfrac{5}{2\pi}\bigg)\quad \longrightarrow \quad \large\boxed{x=\dfrac{5}{8}}}\\\\\\ \dfrac{2\pi}{5}x=\dfrac{7\pi}{4}\quad \longrightarrow \quad \bigg(\dfrac{5}{2\pi}\bigg)\dfrac{2\pi}{5}x=\dfrac{7\pi}{4}\bigg(\dfrac{5}{2\pi}\bigg)\quad \longrightarrow \quad \large\boxed{x=\dfrac{35}{8}}}\\[/tex]
[tex]\dfrac{2\pi}{5}x=\dfrac{\pi}{4}+2\pi\quad \longrightarrow \quad \bigg(\dfrac{5}{2\pi}\bigg)\dfrac{2\pi}{5}x=\dfrac{9\pi}{4}\bigg(\dfrac{5}{2\pi}\bigg)\quad \longrightarrow \quad \large\boxed{x=\dfrac{45}{8}}}\\\\\\ \dfrac{2\pi}{5}x=\dfrac{7\pi}{4}+2\pi\quad \longrightarrow \quad \bigg(\dfrac{5}{2\pi}\bigg)\dfrac{2\pi}{5}x=\dfrac{15\pi}{4}\bigg(\dfrac{5}{2\pi}\bigg)\quad \longrightarrow \quad \large\boxed{x=\dfrac{75}{8}}}\\[/tex]
Notice that inputting n = 2 results in x > 10 so these are the only valid solutions to the equation.
