Question Number 221787 by efronzo1 last updated on 10/Jun/25
$$\:\:\sqrt{\frac{\mathrm{1}−\mathrm{4x}\sqrt{\mathrm{1}−\mathrm{4x}^{\mathrm{2}} }}{\mathrm{2}}}\:=\:\mathrm{1}−\mathrm{8x}^{\mathrm{2}} \\ $$$$\:\:\mathrm{x}=?\: \\ $$
Answered by mr W last updated on 11/Jun/25
$$\mathrm{1}−\mathrm{4}{x}^{\mathrm{2}} =\mathrm{1}−\left(\mathrm{2}{x}\right)^{\mathrm{2}} \geqslant\mathrm{0}\: \\ $$$$\Rightarrow−\mathrm{1}\leqslant\mathrm{2}{x}\leqslant\mathrm{1} \\ $$$${let}\:\mathrm{2}{x}=\mathrm{sin}\:\theta\:{with}\:−\frac{\pi}{\mathrm{2}}\leqslant\theta\leqslant\frac{\pi}{\mathrm{2}} \\ $$$$\sqrt{\frac{\mathrm{1}−\mathrm{2}\:\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta}{\mathrm{2}}}=\mathrm{1}−\mathrm{2}\:\mathrm{sin}^{\mathrm{2}} \:\theta \\ $$$$\sqrt{\frac{\mathrm{1}−\mathrm{sin}\:\mathrm{2}\theta}{\mathrm{2}}}=\mathrm{cos}\:\mathrm{2}\theta \\ $$$$\mathrm{1}−\mathrm{sin}\:\mathrm{2}\theta=\mathrm{2}\:\mathrm{cos}^{\mathrm{2}} \:\mathrm{2}\theta \\ $$$$−\mathrm{sin}\:\mathrm{2}\theta=\mathrm{cos}\:\mathrm{4}\theta \\ $$$$\mathrm{sin}\:\left(−\mathrm{2}\theta\right)=\mathrm{cos}\:\mathrm{4}\theta=\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}−\mathrm{4}\theta\right) \\ $$$$−\mathrm{2}\theta=\frac{\pi}{\mathrm{2}}−\mathrm{4}\theta\:{or}\:\pi+\mathrm{2}\theta=\frac{\pi}{\mathrm{2}}−\mathrm{4}\theta \\ $$$$\Rightarrow\theta=\frac{\pi}{\mathrm{4}}\:{or}\:\theta=−\frac{\pi}{\mathrm{12}} \\ $$$$\Rightarrow{x}=\frac{\mathrm{sin}\:\theta}{\mathrm{2}}=\frac{\mathrm{sin}\:\frac{\pi}{\mathrm{4}}}{\mathrm{2}}=\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\:\checkmark \\ $$$${or} \\ $$$$\Rightarrow{x}=\frac{\mathrm{sin}\:\left(−\frac{\pi}{\mathrm{12}}\right)}{\mathrm{2}}=\frac{\sqrt{\mathrm{2}}−\sqrt{\mathrm{6}}}{\mathrm{8}}\:\checkmark \\ $$