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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!
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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mathmate
 one year ago
Best ResponseYou've already chosen the best response.0dw:1433470200364:dw

manish00333
 one year ago
Best ResponseYou've already chosen the best response.0how would you do it algebraically?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Look 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.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Understand manish? :o 9pi4 = pi/4 + 2pi

manish00333
 one year ago
Best ResponseYou've already chosen the best response.0yeap, 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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