| Resolviendo: |
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Convirtiendo $radianes$ a $sexagesimales$ |
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Se sabe que: $\pi=180^{\circ}$ |
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Reemplazando en el expresión: |
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$\sqrt[\textstyle 3]{5\cot^2{\dfrac{\pi rad}{6}}+3\sec^2{\dfrac{\pi rad}{3}}}$ |
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$\sqrt[\textstyle 3]{5\cot^2{\dfrac{180^{\circ}}{6}}+3\sec^2{\dfrac{180^{\circ}}{3}}}$ |
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$\sqrt[\textstyle 3]{5\cot^2{30^{\circ}}+3\sec^2{60^{\circ}}}$ |
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Recurrimos al triángulo notable: |
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$cot \,{30^{\circ}}=\dfrac{cateto\,adyacente}{cateto\,opuesto}=\dfrac{\sqrt 3}{1}=\sqrt 3$ |
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$sec \,{60^{\circ}}=\dfrac{hipotenusa}{cateto\,adyacente}=\dfrac{2}{1}=2$ |
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Reemplazamos en la expresión: |
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$\sqrt[\textstyle 3]{5\cot^2{30^{\circ}}+3\sec^2{60^{\circ}}}$ |
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$\sqrt[\textstyle 3]{5 \left(\sqrt 3 \right)^{\textstyle 2}+3\left(2 \right)^{\textstyle 2}}$ |
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$\sqrt[\textstyle 3]{5 \left(3 \right)+3\left(4 \right)}$ |
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$\sqrt[\textstyle 3]{15 +12}$ |
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$\sqrt[\textstyle 3]{27}$ |
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$3$ |
| Respuesta: |
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La solución es la Alternativa C |
