Question Number 223673 by gregori last updated on 02/Aug/25
Answered by som(math1967) last updated on 02/Aug/25
$$\:{tan}^{\mathrm{2}} {x}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\:\frac{{sin}^{\mathrm{8}} {x}}{\mathrm{8}}+\frac{\mathrm{cos}\:^{\mathrm{8}} {x}}{\mathrm{27}}=\frac{\mathrm{1}}{\mathrm{125}} \\ $$$$\:\frac{{sin}^{\mathrm{4}} {x}}{\mathrm{2}}\:+\frac{\left(\mathrm{1}−{sin}^{\mathrm{2}} {x}\right)^{\mathrm{2}} }{\mathrm{3}}=\frac{\mathrm{1}}{\mathrm{5}}\: \\ $$$$\Rightarrow\mathrm{3}{sin}^{\mathrm{4}} {x}+\mathrm{2}{sin}^{\mathrm{4}} {x}−\mathrm{4}{sin}^{\mathrm{2}} {x}+\mathrm{2}=\frac{\mathrm{6}}{\mathrm{5}} \\ $$$$\Rightarrow\mathrm{25}{sin}^{\mathrm{4}} {x}−\mathrm{20}{sin}^{\mathrm{2}} {x}+\mathrm{4}=\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{5}{sin}^{\mathrm{2}} {x}−\mathrm{2}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\:\therefore{sin}^{\mathrm{2}} {x}=\frac{\mathrm{2}}{\mathrm{5}}\Rightarrow{cos}^{\mathrm{2}} {x}=\frac{\mathrm{3}}{\mathrm{5}} \\ $$$$\:{sin}^{\mathrm{8}} {x}=\frac{\mathrm{2}^{\mathrm{4}} }{\mathrm{5}^{\mathrm{4}} }\:\:\:\:\Rightarrow{cos}^{\mathrm{8}} {x}=\frac{\mathrm{3}^{\mathrm{4}} }{\mathrm{5}^{\mathrm{4}} } \\ $$$$\therefore\:\frac{{sin}^{\mathrm{8}} {x}}{\mathrm{8}}\:+\frac{{cos}^{\mathrm{8}} {x}}{\mathrm{27}} \\ $$$$=\frac{\mathrm{2}^{\mathrm{4}} }{\mathrm{8}×\mathrm{5}^{\mathrm{4}} }\:+\frac{\mathrm{3}^{\mathrm{4}} }{\mathrm{27}×\mathrm{5}^{\mathrm{4}} }=\frac{\mathrm{2}+\mathrm{3}}{\mathrm{5}^{\mathrm{4}} }=\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{3}} }=\frac{\mathrm{1}}{\mathrm{125}} \\ $$