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kb6

  • 2 years ago

sin (2x) =-1/2

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  1. Mertsj
    • 2 years ago
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    cool

  2. jdoe0001
    • 2 years ago
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    \(\bf sin (2x) =-\cfrac{1}{2}\quad \textit{take }sin^{-1} \\ \quad \\ sin^{-1}[sin ({\color{red}{ 2x}})]=sin^{-1}\left(-\frac{1}{2}\right)\implies {\color{red}{ 2x}}=sin^{-1}\left(-\frac{1}{2}\right)\) solve for "x"

  3. kb6
    • 2 years ago
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    It says to solve the equation on the interval 0<x<2pi

  4. jdoe0001
    • 2 years ago
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    yes so... what did you get for "x"?

  5. surjithayer
    • 2 years ago
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    \[\sin 2x=-\frac{ 1 }{ 2 }=-\sin \frac{ \pi }{6 }=\sin \left( \pi+\frac{ \pi }{6 } \right),\sin \left( 2\pi-\frac{ \pi }{ 6 } \right),\sin \left( 2\pi+\pi+\frac{ \pi }{ 6 } \right),\] \[\sin \left( 2\pi+2\pi-\frac{ \pi }{6 } \right)\] 2x=? x=?

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