Trigo

\hspace{-16}$Using $\bf{Cauchy–Schwarz}$ Inequality::\\\\\\ $\bf{\frac{a^2}{x}+\frac{b^2}{y}\geq \frac{(a+b)^2}{x+y}.}$ and equality hold when $\bf{\frac{a}{x}=\frac{b}{y}}$\\\\\\ Here Given $\bf{\frac{\left(\sin^2x\right)}{2}+\frac{\left(\cos^2x\right)}{3}\geq \frac{\left(\sin^2 x +\cos^2 x\right)}{2+3}}$\\\\\\ Now equality hold when $\bf{\frac{\left(\sin^2 x\right)}{2} = \frac{\left(\cos^2 x\right)}{3}}$\\\\\\ So $\bf{\frac{\left(\sin^2 x\right)}{2} = \frac{\left(\cos^2 x\right)}{3}=\frac{\sin^2 x+\cos^2 x}{2+3}}$\\\\\\ Using Ratio and Proportion:\\\\\\ $\bf{\frac{a}{b}=\frac{c}{d}=\frac{a+c}{b+d}}$\\\\\\ So $\boxed{\bf{\tan^2 x = \frac{2}{3}}}$ and $\bf{\sin^2 x = \frac{2}{5}}$ and $\bf{\cos^2 x = \frac{3}{5}}$\\\\\\ So $\bf{\sin^8 x = \frac{16}{625}\Leftrightarrow \frac{\sin^8 x}{8}=\frac{2}{625}}$ \\\\\\ $\bf{\cos^8 x = \frac{81}{625}\Leftrightarrow \frac{\cos^8 x}{27}=\frac{3}{625}}$\\\\\\ So $\boxed{\bf{\frac{\sin^8 x}{8}+\frac{\cos^8 x}{27}=\frac{2+3}{625}=\frac{1}{125}}}$