Question Number 60814 by Forkum Michael Choungong last updated on 26/May/19
$${find}\:{x}\:{given}\:{that} \\ $$$$\mathrm{9}^{{sin}^{\mathrm{2}} {x}} +\mathrm{9}^{{cos}^{\mathrm{2}} {x}} =\mathrm{2}\: \\ $$$$ \\ $$
Answered by $@ty@m last updated on 26/May/19
$$\mathrm{9}^{{y}} +\mathrm{9}^{\mathrm{1}−{y}} =\mathrm{2} \\ $$$$\mathrm{9}^{{y}} +\frac{\mathrm{9}}{\mathrm{9}^{{y}} }=\mathrm{2} \\ $$$${z}+\frac{\mathrm{9}}{{z}}=\mathrm{2} \\ $$$${z}^{\mathrm{2}} −\mathrm{2}{z}+\mathrm{9}=\mathrm{0} \\ $$$${z}=\frac{\mathrm{2}\pm\sqrt{\mathrm{4}−\mathrm{36}}}{\mathrm{2}} \\ $$$${z}=\frac{\mathrm{2}\pm\mathrm{4}\sqrt{\mathrm{2}}{i}}{\mathrm{2}} \\ $$$${z}=\mathrm{1}\pm\mathrm{2}\sqrt{\mathrm{2}}{i} \\ $$$$\mathrm{9}^{\mathrm{sin}\:^{\mathrm{2}} {x}} =\mathrm{1}\pm\mathrm{2}\sqrt{\mathrm{2}}{i} \\ $$