Să se rezolve ecuatia 3^x+3^-x=10/3

3^x+3^(-x)=10/3
3^x+1/3^x=10/3
3^x=y   y>0
y+1./y=10/3
y²+1=10y/3
3y²-10y+3=0
y1=1/3=>   3^x=1/3 =>   x1=-1
y2=3    3^x=3    x2=1

[tex] 3^{x}+ 3^{-x}= \frac{10}{3} \\ fie:3^x=t;t\ \textgreater \ 0 \\ t+ \frac{1}{t}= \frac{10}{3} \\ t^{2}- \frac{10t}{3}+1=0 \\ 3t^2-10t+3=0 \\ [/tex]
Δ=100-36=64
[tex] t_{1}= \frac{10-8}{6}= \frac{1}{3} \\ t_{2}= \frac{10+8}{6}=3 [/tex]
revenim la substitutie
[tex] 3^{x}= \frac{1}{3};3^x= 3^{-1};x=-1 \\ 3^x=3;x=1 [/tex]
S={-1;1}