-pi(sinpix)= 0
What two ansers should i get? i got x=0 for sinpix and -pi will never equal zero
2007-10-07 07:32:08 · 9 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
π is not a variable. You don't solve for it. Only the x
sin πx=0
πx=0,π,2π,3π... and their negatives
x=0,1,2,3...
2007-10-07 07:37:01 · answer #1 · answered by chasrmck 6 · 0⤊ 0⤋
All you need to concentrate on is
sin(pi * x) = 0
It's correct that x = 0 is one answer but you also have when x = 1 since sin(pi) = 0.
So the answer is x = 0 and 1.
2007-10-07 07:36:45 · answer #2 · answered by clawcomp2 1 · 0⤊ 0⤋
Sin(pi) = 0 so sin(pi x) = 0 when x = 1
2007-10-07 07:36:39 · answer #3 · answered by norman 7 · 0⤊ 0⤋
You don't have 2 answers. pi is a constant coefficient, so you don't solve for it. The solution is for x, and would x= 0,|1|,|2|,etc.
2007-10-07 07:39:04 · answer #4 · answered by cattbarf 7 · 0⤊ 0⤋
-pi(sinpix)= 0
=>(sinpix)= 0/ -pi
=>(sinpix)= 0
=> pix= 0,pi.2pi.npi
=> x= 0,pi/pi. 2pi/pi. npi/pi
=> x= 0, 1,2,3,...n
2007-10-07 07:39:05 · answer #5 · answered by harry m 6 · 0⤊ 0⤋
substitute pi for 180
2007-10-07 07:36:59 · answer #6 · answered by Anonymous · 0⤊ 0⤋
-pi=0(impossible)
or
sin(pi*x)=0
sin(pi*x)=sin(0)
pi*x=2kpi+0 or pi*x=2kp+p
x=2k or x=2k+1
These are the answers!So basically, x has to be an integer!
2007-10-07 07:41:35 · answer #7 · answered by Santa's little helper 2 · 0⤊ 0⤋
placed one million - cos^2 in decision to sin^2, then permit m = cos x. this might leave you with the quadratic equation m^2 + 4m - one million = 0. resolve for m then placed cos(x) lower back in place and resolve for x.
2017-01-03 06:20:06 · answer #8 · answered by ? 4 · 0⤊ 0⤋
what??
2007-10-07 07:34:26 · answer #9 · answered by serser92695 2 · 0⤊ 1⤋