Question Number 90507 by student work last updated on 24/Apr/20
$$ \\ $$$$\:\mathrm{if}\:\:\mathrm{2}^{\mathrm{sin}\:\mathrm{x}} +\mathrm{2}^{\mathrm{cos}\:\mathrm{x}} =\mathrm{2}^{\mathrm{1}+\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}} \:\:\mathrm{then}\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{x}=? \\ $$
Commented by student work last updated on 24/Apr/20
$$\mathrm{who}\:\mathrm{can}\:\mathrm{solve}? \\ $$
Commented by MJS last updated on 24/Apr/20
$${x}=\frac{\pi}{\mathrm{4}}+\mathrm{2}{n}\pi;\:{n}\in\mathbb{Z} \\ $$$$\mathrm{sin}\:\frac{\pi}{\mathrm{4}}\:=\mathrm{cos}\:\frac{\pi}{\mathrm{4}}\:=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$$\mathrm{2}^{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}} +\mathrm{2}^{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}} =\mathrm{2}×\mathrm{2}^{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}} =\mathrm{2}^{\mathrm{1}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}} =\mathrm{2}^{\mathrm{1}+\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}} \\ $$