My Math Forum Don't know what to do now

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 March 12th, 2017, 05:35 AM #1 Newbie   Joined: Feb 2017 From: Brazil Posts: 5 Thanks: 0 Don't know what to do now There is this equation: Î˜(T)=V0Sin(2T)+Î˜0Cos(2T) It wants the time (T) where Î˜ will be 0. And say V0 and Î˜0 has same value. So I did this: V0 and Î˜0 = X 0=Xsin(2t)+XCos(2t) So I stop here: 0=X(Sin2T+Cos2T) It also says to consider PI as 22/7. Last edited by skipjack; March 12th, 2017 at 05:58 AM.
 March 12th, 2017, 05:57 AM #2 Global Moderator   Joined: Dec 2006 Posts: 16,775 Thanks: 1235 sin(2T) + cos(2T) â‰¡ âˆš2sin(2T + $\pi$/4)
March 12th, 2017, 06:43 AM   #3
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Quote:
 Originally Posted by skipjack sin(2T) + cos(2T) â‰¡ âˆš2sin(2T + $\pi$/4)
Sorry, still don't know what to do.

0=Sqrt(2)XSin(2T+ $\pi/$4)

0=Sqrt(2)XSin(2T+ (88/7))

 March 12th, 2017, 08:51 AM #4 Global Moderator   Joined: Dec 2006 Posts: 16,775 Thanks: 1235 $\pi/4$ isn't 88/7. sin(2T + $\pi$/4) = 0 implies 2T + $\pi$/4 = k$\pi$, where k is an integer, so T = (k/2 - 1/8)$\pi$.
March 12th, 2017, 09:06 AM   #5
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Quote:
 Originally Posted by skipjack $\pi/4$ isn't 88/7. sin(2T + $\pi$/4) = 0 implies 2T + $\pi$/4 = k$\pi$, where k is an integer, so T = (k/2 - 1/$\pi$.
I see, but the exercise requires a number as an answer, so there might be a way to eliminate de K. I said $\pi$4 is 88/7 because the exercise says to consider $\pi$ as 22/7

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