Odpowiedź:
[tex]\huge\boxed{x=\dfrac{\pi}{2}+k\pi\ \vee\ x=\dfrac{\pi}{4}+k\pi,\ k\in\mathbb{C}}[/tex]
Szczegółowe wyjaśnienie:
[tex]\cos^3x+\sin^3x=\sin x\qquad|a^3+b^3=(a+b)(a^2-ab+b^2)\\\\(\cos x+\sin x)(\cos^2x-\sin x\cos x+\sin^2x)=\sin x\qquad|\sin^2x+\cos^2x=1\\\\(\cos x+\sin x)(1-\sin x\cos x)=\sin x\\\\\cos x-\sin x\cos^2x+\sin x-\sin^2x\cos x=\sin x\qquad|-\sin x\\\\\cos x-\sin x\cos^2x-\sin^2x\cos x=0\\\\\cos x(1-\sin x\cos x-sin^2x)=0\iff\cos x=0\ \vee\ 1-\sin x\cos x-\sin^2x=0\\\\\sin^2x+\cos^2x-\sin x\cos x-\sin^2x=0\\\\\cos^2x-\sin x\cos x=0\\\\\cos x(\cos x-\sin x)=0\iff\cos x=0\ \vee\ \cos x-\sin x=0\\\\\cos x=\sin x[/tex]
[tex]\cos x=0\Rightarrow x=\dfrac{\pi}{2} +k\pi,\ k\in\mathbb{C}\\\\\cos x=\sin x\Rightarrow x=\dfrac{\pi}{4}+k\pi,\ k\in\mathbb{C}[/tex]
