Question Number 175369 by infinityaction last updated on 28/Aug/22
$$\:\:\mathrm{solve}\:\mathrm{for}\:{x} \\ $$$$\:\:\:\mathrm{log}_{\mid\mathrm{sin}{x}\mid\:} \left({x}^{\mathrm{2}} −\mathrm{8}{x}+\mathrm{23}\right)\:>\:\frac{\mathrm{3}}{\mathrm{log}_{\mathrm{2}} \mid\mathrm{sin}{x}\mid\:} \\ $$
Answered by floor(10²Eta[1]) last updated on 28/Aug/22
$$\frac{\mathrm{log}\left(\mathrm{x}^{\mathrm{2}} −\mathrm{8x}+\mathrm{23}\right)}{\mathrm{log}\mid\mathrm{sinx}\mid}>\frac{\mathrm{3log2}}{\mathrm{log}\mid\mathrm{sinx}\mid} \\ $$$$\mathrm{log}\left(\mathrm{x}^{\mathrm{2}} −\mathrm{8x}+\mathrm{23}\right)>\mathrm{log8} \\ $$$$\mathrm{x}^{\mathrm{2}} −\mathrm{8x}+\mathrm{23}>\mathrm{8}\Rightarrow\mathrm{x}^{\mathrm{2}} −\mathrm{8x}+\mathrm{15}>\mathrm{0} \\ $$$$\mathrm{x}<\mathrm{3}\:\mathrm{or}\:\mathrm{x}>\mathrm{5} \\ $$$$\mathrm{sinx}\neq\mathrm{0}\Rightarrow\mathrm{x}\neq\mathrm{k}\pi,\:\mathrm{k}\in\mathbb{Z} \\ $$$$\mathrm{log}\mid\mathrm{sinx}\mid\neq\mathrm{0}\Rightarrow\mid\mathrm{sinx}\mid\neq\mathrm{1}\Rightarrow\mathrm{x}=\frac{\left(\mathrm{2n}+\mathrm{1}\right)\pi}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{S}=\left\{\mathrm{x}\in\mathbb{R}−\left\{\mathrm{k}\pi,\:\frac{\left(\mathrm{2n}+\mathrm{1}\right)\pi}{\mathrm{2}}\right\},\:\mathrm{k},\mathrm{n}\in\mathbb{Z}\mid\mathrm{x}\in\left(−\infty,\mathrm{3}\right)\cup\left(\mathrm{5},\infty\right)\right\} \\ $$
Commented by infinityaction last updated on 28/Aug/22
$${check}\:{your}\:{solution} \\ $$