Question Number 132107 by weltr last updated on 11/Feb/21
$$\left({arcsin}\mid\mathrm{ln}^{\mathrm{3}} \left({cot}\mathrm{4}\:−\:{x}\right)\mid\:+\:\mathrm{4}!\right)^{\frac{\mathrm{1}}{\mathrm{19}}} \:=\:\mathrm{5} \\ $$
Answered by Olaf last updated on 11/Feb/21
$$\left(\mathrm{arcsin}\mid\mathrm{ln}^{\mathrm{3}} \left(\mathrm{cot4}−{x}\right)\mid+\mathrm{4}!\right)^{\frac{\mathrm{1}}{\mathrm{19}}} \:=\:\mathrm{5} \\ $$$$\mathrm{arcsin}\mid\mathrm{ln}^{\mathrm{3}} \left(\mathrm{cot4}−{x}\right)\mid\:=\:\mathrm{5}^{\mathrm{19}} −\mathrm{24}\:>\:\frac{\pi}{\mathrm{2}} \\ $$$$\mathcal{S}_{\mathbb{R}} \:=\:\emptyset \\ $$