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ksaimouli

  • 2 years ago

simplyfy

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  1. ksaimouli
    • 2 years ago
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    \[0.2=\cos(\frac{ \pi t }{ 12 })\]

  2. ksaimouli
    • 2 years ago
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    solve for t

  3. Jonask
    • 2 years ago
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    \[\frac{\pi t}{12}=\arccos (0.2)\]

  4. ksaimouli
    • 2 years ago
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    oopss i did not think this way thx

  5. Jonask
    • 2 years ago
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    but are you allowed to use calculator \[\pi t/12=\pm \arccos(0.2)+2 \pi k,\forall k\in Z\]

  6. ksaimouli
    • 2 years ago
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    i got 5.231 but i only got one what about other value

  7. ksaimouli
    • 2 years ago
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    |dw:1359586504389:dw|

  8. ksaimouli
    • 2 years ago
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    @Jonask

  9. Jonask
    • 2 years ago
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    notice that there are infinite number of solutiond but the general solution of these is \[\huge t=\frac{12}{\pi}\arccos(0.2)+2\pi k\] where k can be chosen from interger k={....-2,-1,0,1,2,3,4...}

  10. ksaimouli
    • 2 years ago
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    where that +2 pi k came from

  11. Jonask
    • 2 years ago
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    this is for every period of cos wich is every 360 =2pi but if you never did it just consider your answer + and -5.321 and 360-5.321

  12. ksaimouli
    • 2 years ago
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    but the answers are +5.321 and 18.767

  13. Jonask
    • 2 years ago
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    first \[\frac{t \pi}{12}=\cos^{-1}{0.2}=78\]

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