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Question Number 164050 by mathlove last updated on 13/Jan/22

Answered by Ar Brandon last updated on 13/Jan/22

$$x-\frac{1}{8}>0\ne 1\Rightarrow x>\frac{1}{8}\ne \frac{9}{8}....\left(1\right)$$$$x^2-\frac{1}{9}>0\Rightarrow x<-\frac{1}{3}\cup x>\frac{1}{3}...\left(2\right)$$$${\sin }^{-1}(2-x)>0\Rightarrow 2-x>0\Rightarrow x<2...\left(3\right)$$$$\left(1\right),\left(2\right),\left(3\right)$$$$\Rightarrow \mathrm{S}=\{x:\frac{1}{3}<x<\frac{9}{8}\cup \frac{9}{8}<x<2\}$$

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