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  • 18-08-2022
  • Mathematics
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What is the solution to the trigonometric inequality 2-3csc(x) > 8 over the interval radians?

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subtomex0
subtomex0 subtomex0
  • 18-08-2022

[tex]2 - 3csc(x) > 8 \\ 2 - \frac{3}{sin(x)} > 8 \\ - 6 > \frac{3}{sin(x)} \\ - 2 > \frac{1}{sin(x)} \\ \frac{ - 1}{2} < sin(x) \: \: or \: \: \: sin(x) > \frac{ - 1}{2} \\ \\ [/tex]

[tex]sin( \frac{ - \pi}{6} ) = \frac{ - 1}{2} [/tex]

[tex]x \: in \: \: [0, \frac{7\pi}{6}[U] \frac{11 \pi }{6} ,2\pi] + 2k\pi[/tex]

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