bellafielder bellafielder
  • 23-06-2022
  • Mathematics
contestada

Please help! Solve the equation on the interval [0,2pi)

Please help Solve the equation on the interval 02pi class=

Respuesta :

megank14699 megank14699
  • 23-06-2022

Answer:

[tex]\frac{5pi}{6}[/tex]

Step-by-step explanation:

Answer Link
semsee45
semsee45 semsee45
  • 23-06-2022

Answer:

[tex]x = \dfrac{\pi}{6}, \dfrac{5 \pi}{6}[/tex]

Step-by-step explanation:

[tex]\begin{aligned}3 \sin x & = \sin x + 1\\\implies 3 \sin x - \sin x & = 1\\2 \sin x & = 1\\\sin x & = \dfrac{1}{2}\\\implies x & = \sin^{-1}\left(\dfrac{1}{2}\right)\\x & = \dfrac{\pi}{6} \pm 2 \pi n, \dfrac{5 \pi}{6}\pm 2 \pi n\end{aligned}[/tex]

Therefore, the value of x in the interval [0, 2π) is:

[tex]x = \dfrac{\pi}{6}, \dfrac{\boxed{5} \pi}{\boxed{6}}[/tex]

Answer Link

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