Question Number 146865 by mathdanisur last updated on 16/Jul/21
$${arcsin}\left({x}^{\mathrm{2}} −\mathrm{4}\right)\:=\:{arcsin}\left(\mathrm{2}{x}\:+\:\mathrm{4}\right) \\ $$$$\Rightarrow\:{x}\:=\:? \\ $$
Answered by Olaf_Thorendsen last updated on 16/Jul/21
![arcsin(x^2 −4) = arcsin(2x+4) ⇔ { ((−1 ≤x^2 −4 ≤ +1)),((−1 ≤ 2x+4 ≤ +1)),((x^2 −4 = 2x+4)) :} ⇔ { ((x∈[−(√5);−(√3)]∪[+(√3);+(√5)])),((x∈[−(5/2);−(3/2)])),((x^2 −2x−8 = 0)) :} x^2 −2x−8 = (x+2)(x−4) = 0 x = −2 or x = 4 but 4∉[−(5/2);−(3/2)] S = {−2}](https://www.tinkutara.com/question/Q146869.png)
$$\mathrm{arcsin}\left({x}^{\mathrm{2}} −\mathrm{4}\right)\:=\:\mathrm{arcsin}\left(\mathrm{2}{x}+\mathrm{4}\right) \\ $$$$\Leftrightarrow\:\begin{cases}{−\mathrm{1}\:\leqslant{x}^{\mathrm{2}} −\mathrm{4}\:\leqslant\:+\mathrm{1}}\\{−\mathrm{1}\:\leqslant\:\mathrm{2}{x}+\mathrm{4}\:\leqslant\:+\mathrm{1}}\\{{x}^{\mathrm{2}} −\mathrm{4}\:=\:\mathrm{2}{x}+\mathrm{4}}\end{cases} \\ $$$$\Leftrightarrow\:\begin{cases}{{x}\in\left[−\sqrt{\mathrm{5}};−\sqrt{\mathrm{3}}\right]\cup\left[+\sqrt{\mathrm{3}};+\sqrt{\mathrm{5}}\right]}\\{{x}\in\left[−\frac{\mathrm{5}}{\mathrm{2}};−\frac{\mathrm{3}}{\mathrm{2}}\right]}\\{{x}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{8}\:=\:\mathrm{0}}\end{cases} \\ $$$${x}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{8}\:=\:\left({x}+\mathrm{2}\right)\left({x}−\mathrm{4}\right)\:=\:\mathrm{0} \\ $$$${x}\:=\:−\mathrm{2}\:\mathrm{or}\:{x}\:=\:\mathrm{4}\:\mathrm{but}\:\mathrm{4}\notin\left[−\frac{\mathrm{5}}{\mathrm{2}};−\frac{\mathrm{3}}{\mathrm{2}}\right] \\ $$$$\mathcal{S}\:=\:\left\{−\mathrm{2}\right\} \\ $$
Commented by mathdanisur last updated on 16/Jul/21
$${thankyou}\:{Ser}\:{cool} \\ $$