Rozwiązanie:
[tex]8\sin^{4}x-\sin x+8\sin ^{3}x \cos x-\cos x=0[/tex]
[tex]\sin x(8 \sin^{3}x-1)+\cos x(8 \sin^{3}x-1)=0[/tex]
[tex](8\sin^{3}x-1)(\sin x +\cos x)=0[/tex]
[tex](2\sin x-1)(\sin^{2}x+\sin x+1)(\sin x + \cos x) =0[/tex]
Zauważmy, że [tex]\sin ^{2}x+\sin x+1>0[/tex] dla [tex]x \in \mathbb{R}[/tex]. Zatem:
[tex]2 \sin x -1 =0 \vee \sin x +\cos x = 0[/tex]
[tex]$\sin x=\frac{1}{2} \vee \sqrt{2} \sin \Big(x+\frac{\pi}{4} \Big)=0[/tex]
[tex]$x=\frac{\pi}{6} +2k\pi \vee x=\frac{5\pi}{6} +2k\pi \vee x+\frac{\pi}{4} =k\pi[/tex]
[tex]$x=\frac{\pi}{6} +2k\pi \vee x=\frac{5\pi}{6} +2k\pi \vee x =-\frac{\pi}{4}+k\pi[/tex]
[tex]k \in \mathbb{Z}[/tex]
