Question Number 139141 by bramlexs22 last updated on 23/Apr/21
$$\:\:\:\mathrm{3}^{\mathrm{sin}\:\mathrm{x}} \:\mathrm{cos}\:\mathrm{x}\:−\mathrm{3}^{\mathrm{cos}\:\mathrm{x}} \:\mathrm{sin}\:\mathrm{x}\:=\:\mathrm{0} \\ $$
Answered by phanphuoc last updated on 23/Apr/21
$$\frac{\mathrm{3}^{{sinx}} }{{sinx}}=\frac{\mathrm{3}^{{cosx}} }{{cosx}}\rightarrow{sinx}={cosx}\rightarrow{tgx}=\mathrm{1}\rightarrow{x}=\pi/\mathrm{4}+{k}\pi \\ $$$${sinxcosx}=\mathrm{0}\rightarrow{sin}\mathrm{2}{x}=\mathrm{0}\rightarrow\mathrm{2}{x}={k}\pi\rightarrow{x}={k}\pi/\mathrm{2} \\ $$
Commented by mr W last updated on 23/Apr/21
$${x}\neq\frac{{k}\pi}{\mathrm{2}} \\ $$
Answered by EDWIN88 last updated on 23/Apr/21
$$\mathrm{x}\:=\:\frac{\pi}{\mathrm{4}}\&\mathrm{x}=\frac{\mathrm{5}\pi}{\mathrm{4}}\: \\ $$
Commented by mr W last updated on 23/Apr/21
$${i}\:{think}\:{generally}\:{x}={k}\pi+\frac{\pi}{\mathrm{4}} \\ $$