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Question Number 132107 by weltr last updated on 11/Feb/21

$${(arcsin\mid {\ln }^3(cot4-x)\mid +4!)}^{\frac{1}{19}}=5$$

Answered by Olaf last updated on 11/Feb/21

$${(\arcsin \mid {\ln }^3(\cot 4-x)\mid +4!)}^{\frac{1}{19}}=5$$$$\arcsin \mid {\ln }^3(\cot 4-x)\mid =5^{19}-24>\frac{\pi }{2}$$$${\mathcal{S}}_{\mathbb{R}}=\mathrm{\varnothing }$$

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