Przekształćmy równanie:

${\sin }^{6}x+{\cos }^{6}x=k$  

${\left({\sin }^{2}x\right)}^3+{\left({\cos }^{2}x\right)}^3=k$  

Ze wzoru na sumę sześcianów:

$\left({\sin }^{2}x+{\cos }^{2}x\right)\left({\left({\sin }^{2}x\right)}^2-{\sin }^{2}x{\cos }^{2}x+{\left({\cos }^{2}x\right)}^2\right)=k$  

$1\cdot \left({\sin }^{4}x-{\sin }^{2}x{\cos }^{2}x+{\cos }^{4}x\right)=k$  

${\left({\sin }^{2}x+{\cos }^{2}x\right)}^2-2{\sin }^{2}x{\cos }^{2}x-{\sin }^{2}x{\cos }^{2}x=k$  

${\left({\sin }^{2}x+{\cos }^{2}x\right)}^2-3{\sin }^{2}x{\cos }^{2}x=k$  

$1-\frac{3}{4}\cdot 4{\sin }^{2}x{\cos }^{2}x=k$  

$1-\frac{3}{4}\cdot {\left(2\sin x\cos x\right)}^2=k$  

$1-\frac{3}{4}{\left(\sin 2x\right)}^2=k$  

$-\frac{3}{4}{\left(\sin 2x\right)}^2=k-1$  

${\left(\sin 2x\right)}^2=-\frac{4}{3}k+\frac{4}{3}$  

${\left(\sin 2x\right)}^2=-\frac{4}{3}\left(k-1\right)$  

Jeżeli ma istnieć rozwiązanie to:

$k-1\le 0$  

$k\le 1$  

A więc:

$\left|\sin 2x\right|=\sqrt{-\frac{4}{3}\left(k-1\right)}$  

Wiemy, że:

$0\le \left|\sin 2x\right|\le 1$  

a więc:

$0\le \sqrt{-\frac{4}{3}\left(k-1\right)}\le 1\mid {}^2$  

$0\le \left|-\frac{4}{3}\left(k-1\right)\right|\le 1$  

$0\le -\frac{4}{3}\left(k-1\right)\le 1\mid \cdot \left(-\frac{3}{4}\right)$  

$0\ge k-1\ge -\frac{3}{4}$  

$1\ge k\ge \frac{1}{4}$  

A więc:

$k\in \left[\frac{1}{4},1\right]$