Question

FGV – adaptada) A soma das raízes da equação sen2x – sen (–x) = 0, no intervalo [0,2π], é

Escolha uma:
a. 3π/2
b. 9π/2
c. 5π/2
d. 3π
e. 7π/2

Answer 1

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$\boxed{\begin{array}{l}\sf sen(2x)-sen(-x)=0\\\sf sen(-x)=-sen(x)\\\sf sen(2x)-(-sen(x))=0\\\sf sen(2x)+sen(x)=0\\\sf 2sen(x)cos(x)+sen(x)=0\\\sf sen(x)(2cos(x)+1)=0\\\sf sen(x)=0\implies x=0~ou~x=\pi\\\sf 2cos(x)+1=0\\\sf 2cos(x)=-1\\\sf cos(x)=-\dfrac{1}{2}\\\sf x=\dfrac{2\pi}{3}~ou~x=\dfrac{4\pi}{3}\end{array}}$

$\boxed{\begin{array}{l}\underline{\rm verificac_{\!\!,}\tilde ao\!:}\\\sf x=0: \\\sf sen(2\cdot0)-sen(0)=0\checkmark\\\sf x=\pi:\\\sf sen(2\pi)-sen(-\pi)=0-0=0\checkmark\\\sf x=\dfrac{2\pi}{3}:\\\sf sen\bigg(2\cdot\dfrac{2\pi}{3}\bigg)-sen\bigg(-\dfrac{\pi}{3}\bigg)=-\dfrac{\sqrt{3}}{2}+\dfrac{\sqrt{3}}{2}=0\checkmark\\\sf x=\dfrac{4\pi}{3}:\\\sf sen\bigg(2\cdot\dfrac{4\pi}{3}\bigg)-sen\bigg(\dfrac{4\pi}{3}\bigg)=\dfrac{\sqrt{3}}{2}-\dfrac{\sqrt{3}}{2}=0\checkmark\end{array}}$

$\Large\boxed{\begin{array}{l}\sf S=\bigg\{0,\pi,\dfrac{2\pi}{3},\dfrac{4\pi}{3}\bigg\}\\\sf soma=0+\pi+\dfrac{2\pi}{3}+\dfrac{4\pi}{3}\\\sf soma=\dfrac{3\pi}{3}+\dfrac{2\pi}{3}+\dfrac{4\pi}{3}\\\sf soma=\dfrac{9\pi}{3}\\\sf soma=3\pi\\\huge\boxed{\boxed{\boxed{\boxed{\sf\maltese~alternativa~d}}}}\end{array}}$