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Math II question? O.O

MarieMMarieM Registered User Posts: 185 Junior Member
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?
Post edited by MarieM on

Replies to: Math II question? O.O

  • thebigonethebigone Registered User Posts: 80 Junior Member
    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
  • MarieMMarieM Registered User Posts: 185 Junior Member
    I'm dumb.
    I surprise myself at times with it. xDD

    Anyway, thanks thebigone
This discussion has been closed.