Concursos Públicos ⇒ fgv-funções trigonometricas Tópico resolvido

[Pdalindão]

Considere a função 𝑓: [0, 3𝜋/2] → ℝ tal que
𝑓(𝑥) = 𝑠𝑒𝑛 (2𝑥 − 𝜋/3) + 1/2
Chama-se zero da função 𝑓(𝑥) ao valor de 𝑥 que anula 𝑓.
A soma de todos os zeros dessa função é:

(A)5𝜋/6
(B)11𝜋/6
(C)19𝜋/12
(D)21𝜋/12
(E)23𝜋/12

Resposta

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E

[petras]

Pdalindão,

[tex3]\mathsf
{sen(2x-\frac{\pi}{3})=-\frac{1}{2} \implies 2x-\frac{\pi}{3} = \frac{7\pi}{6} \therefore x = \frac{3\pi}{4}\\
2x-\frac{\pi}{3} = \frac{11\pi}{6} \therefore x = \frac{13\pi}{12}\\
2x-\frac{\pi}{3} = -\frac{\pi}{6} \therefore x = \frac{\pi}{12}\\
\frac{3\pi}{4}+\frac{13\pi}{12}+\frac{\pi}{12} = \boxed{\frac{23\pi}{12}}


}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\mathsf
{sen(2x-\frac{\pi}{3})=-\frac{1}{2} \implies 2x-\frac{\pi}{3} = \frac{7\pi}{6} \therefore x = \frac{3\pi}{4}\\
2x-\frac{\pi}{3} = \frac{11\pi}{6} \therefore x = \frac{13\pi}{12}\\
2x-\frac{\pi}{3} = -\frac{\pi}{6} \therefore x = \frac{\pi}{12}\\
\frac{3\pi}{4}+\frac{13\pi}{12}+\frac{\pi}{12} = \boxed{\frac{23\pi}{12}}


}[/tex3]