
Algebra PreAlgebra and Basic Algebra Math Forum 
 LinkBack  Thread Tools  Display Modes 
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: 17,722 Thanks: 1359 
sin(2T) + cos(2T) â‰¡ âˆš2sin(2T + $\pi$/4)

March 12th, 2017, 06:43 AM  #3 
Newbie Joined: Feb 2017 From: Brazil Posts: 5 Thanks: 0  
March 12th, 2017, 08:51 AM  #4 
Global Moderator Joined: Dec 2006 Posts: 17,722 Thanks: 1359 
$\pi/4$ 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 
Newbie Joined: Feb 2017 From: Brazil Posts: 5 Thanks: 0  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
