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Segayo2007
a fost răspuns

Sa se rezolve in multimea nr reale: arctg√3+arctgx=π/2...Vreo idee?

Răspuns :

Roxadria
arctg√3=π/3
π/3+arctgx=π/2
arctgx=π/6
x=√3/3
Aeerdnadeutza
pai arctg√3=π/3
π/3+arctgx=π/2
arctgx=-π/6
S:x∈{arctg-π/6+kπ/k∈R}
S:x∈{-√3/3+kπ/k∈R}

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