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May 14th, 2009, 06:06 AM  #1 
Newbie Joined: Jul 2008 Posts: 12 Thanks: 0  Trigonometric equation doesn’t match the graph
Solutions for 2 cos x= tan2x in the domain 0< x< 2pi Sin2x/cos2x = 2cosx Sin2x = 2cosx.cos2x 2sinx.cosx  2cosx (cos^2x – sin^2x) = 0 Cosx (2sinx – 2(cos^2x –sin^2x) = 0 Cosx (2sinx – 2(1 – sin^2x –sin^2x) = 0 Cosx [(2sinx – 2(1 –2sin^2x)] = 0 Cosx (2sinx + 4sin^2x 2) = 0 2Cosx (sinx +2sin^2x 1) = 0 Cosx = o, x = pi/2, 3pi/2 sinx +2sin^2x 1 = 0 sinx =1/2 , x= pi/6, 5pi/6 sinx = 1, x = pi/2, or 3pi/2 so the solutions are pi/6, 5pi/6, 3pi/2, pi/2 But when we plot 2cosx and tan2x on the same graph they do intersect at six points, and what I am getting are 4 solutions, are there any other solution to this equation. Shouldn’t the points of intersection indicate that there are six possible solutions to 2 cos x= tan2x 
May 14th, 2009, 07:28 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 20,753 Thanks: 2136 
The graphs have the same slope at x = 3pi/2, so they have a single intersection point there instead of the 4th, 5th and 6th intersection points shown in your diagram.

May 14th, 2009, 07:55 AM  #3 
Newbie Joined: Jul 2008 Posts: 12 Thanks: 0  Re: Trigonometric equation doesn’t match the graph
i do agree that the slopes(dy/dx) at 3pi/2 is the same for 2cosx and tanx, but what about the points below and above 3pi/2, aren't they the solutions to 2cosx=tan2x, and if yes then whar are these points

May 14th, 2009, 07:56 AM  #4  
Senior Member Joined: Jul 2008 Posts: 895 Thanks: 0  Re: Trigonometric equation doesn’t match the graph Quote:
[attachment=0:24hvruiu]Trig.jpg[/attachment:24hvruiu]  
May 14th, 2009, 08:21 AM  #5 
Newbie Joined: Jul 2008 Posts: 12 Thanks: 0  Re: Trigonometric equation doesn’t match the graph
thanks dave


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