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anonymous

  • one year ago

Solve: sin^(2) x - Sin x =0, for 0≤x≤3

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  1. Michele_Laino
    • one year ago
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    here we have to change variable, namely: \[\Large \sin x = t\]

  2. Michele_Laino
    • one year ago
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    after that your equation can be rewritten as follows: \[\Large {t^2} - t = 0\]

  3. Michele_Laino
    • one year ago
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    please solve that equation for t

  4. anonymous
    • one year ago
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    So t=0?

  5. Michele_Laino
    • one year ago
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    hint, we can factor out t at left side, so we can write: \[\Large t\left( {t - 1} \right) = 0\]

  6. anonymous
    • one year ago
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    Oh, so t= 0 and 1?

  7. Michele_Laino
    • one year ago
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    that's right!

  8. Michele_Laino
    • one year ago
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    now t=0 means: sin x=0 what is x?

  9. anonymous
    • one year ago
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    x= Radical 3 over 2?

  10. anonymous
    • one year ago
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    Oops!

  11. Michele_Laino
    • one year ago
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    no, the solution of that equation: sin x=0, are: \[x = 0,\pi ,2\pi ,3\pi ...\]

  12. Michele_Laino
    • one year ago
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    nevertheless only x=0, is acceptable, since pi >3

  13. anonymous
    • one year ago
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    Oh, i get it! Thank you!

  14. Michele_Laino
    • one year ago
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    please wait, we have to solve this equation: t=1 which means sin x =1

  15. Michele_Laino
    • one year ago
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    for which values x, we have: sin x= 1?

  16. Michele_Laino
    • one year ago
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    \[\sin x = 1 \Rightarrow x = ...?\]

  17. anonymous
    • one year ago
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    Pie over 2 and 3 pie over 2

  18. Michele_Laino
    • one year ago
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    no, the solutions are: \[x = \frac{\pi }{2},2\pi + \frac{\pi }{2},4\pi + \frac{\pi }{2}...\]

  19. anonymous
    • one year ago
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    Oh yea, my bad.

  20. Michele_Laino
    • one year ago
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    nevertheless only x= pi/2 is acceptable

  21. anonymous
    • one year ago
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    I get it. thanks :D

  22. Michele_Laino
    • one year ago
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    thanks! :)

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