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manish00333

  • one year ago

Solve sin2x= Root2/2 for 0 to 2pi. i got the first two answers: pi/8 and 3pi/8, but i don't understand how to get the other two: 9pi/8 and 11pi/8. thanks!

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  1. mathmate
    • one year ago
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    |dw:1433470200364:dw|

  2. manish00333
    • one year ago
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    how would you do it algebraically?

  3. zepdrix
    • one year ago
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    Look further than 2pi initially. Your first four angles (even though they're not within 2pi) are, pi/4, 3pi/4, 9pi/4, 11pi/4 ya? :) so i went a full rotation around to find the next two. the reason you do this is because those angles WILL end up being within 2pi once you divide by 2 as your final step.

  4. zepdrix
    • one year ago
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    Understand manish? :o 9pi4 = pi/4 + 2pi

  5. manish00333
    • one year ago
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    yeap, thanks! :D So, pi/4 + 2pi = 9pi/4 and 3pi/4 + 2pi = 11pi/4 and when i divide them by 2, i get 9pi/8 and 11pi/8.

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