Răspuns:
x∈(0; π/2), unde sinx>0, tgx>0, ctgx>0
Explicație pas cu pas:
[tex]cosx=\frac{\sqrt{5}}{3},~sin^{2}x+cos^{2}x=1,~atunci~sin^{2}x}+(\frac{\sqrt{5}}{3})^{2}=1\\sin^{2}x=1-\frac{5}{9} =\frac{4}{9}~deci~sinx=\sqrt{\frac{4}{9}}=\frac{2}{3}\\tgx=\frac{sinx}{cosx}=\frac{2}{3}:\frac{\sqrt{5}}{3} =\frac{2}{3}*\frac{3}{\sqrt{5} }=\frac{2}{\sqrt{5} }=\frac{2\sqrt{5}} {5},~ctgx=\frac{cosx}{sinx}=\frac{\sqrt{5}}{2}[/tex]
