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 11-08-2010, 09:28 AM #1 Junior Member   Join Date: Jun 2010 Posts: 175 Math II question? O.O When the graph of y=Sin2x is drawn for all values of x between 10 degrees and 350 degrees, it crosses the x-axis A) zero times B) 1 time C) 2 times D) 3 times E) 6 times Barrons^ Model Test 2, Question 10. The answer is E. and I don't get it. Sin2x=0 when 2x=0, 180, 360, 540, 720.. So..x can be 90, 180, 270, 360, 540, no? Why not? I mean..x should be between 10 and 350 degrees, not 2x, so why can't we include 360 and 540? Can somebody please explain? Reply
 11-08-2010, 12:29 PM #2 Junior Member   Join Date: Aug 2010 Posts: 79 no! first you can just draw the function on your calculator for x between 10 and 350 degrees and count the points or : sin (2x)=0 then 2x= 0 , 180 , 360, 540 , 720... so x = 0, 90, 180, 270, 360... but x is between 10 and 350 degrees so we have three solutions : 90, 180 and 270 Reply
 11-08-2010, 02:48 PM #3 Junior Member   Join Date: Jun 2010 Posts: 175 nevermind. I'm dumb. Seriously. I surprise myself at times with it. xDD Anyway, thanks thebigone Reply

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