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Math II question? O.O

Registered User Posts: 185 Junior Member
edited November 2010
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.

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?

Post edited by MarieM on

Replies to: Math II question? O.O

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

Anyway, thanks thebigone
This discussion has been closed.