Răspuns :

[tex]sin^2x+cos^2x=1\Rightarrow cos^2x=1-\dfrac{2+\sqrt3}{4}=\dfrac{2-\sqrt3}{4}\Rightarrow[/tex]

[tex]\Rightarrow cosx=\dfrac{\sqrt{2-\sqrt3}}{2}[/tex] Am luat semnul + deoarece x este in primul cadran.

[tex]tgx=\dfrac{sinx}{cosx}=\sqrt{\dfrac{2+\sqrt3}{2-\sqrt3}}[/tex] Dupa rationalizarea numitorului

[tex]tgx=2+\sqrt3[/tex]

[tex]tg2x=\dfrac{2tgx}{1-tg^2x}=\dfrac{2(2+\sqrt3)}{1-(2+\sqrt3)^2}=\dfrac{2(2+\sqrt3)}{-6-4\sqrt3}=\dfrac{2(2+\sqrt3)}{-2\sqrt3(\sqrt3+2)}=[/tex]

[tex]=\dfrac{1}{\sqrt3}[/tex]

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