f(x)=4sin[(pi/2)x]+2

Find the first three positive values of x for f(x)=5.

I was able to get one of the answers(0.54) and another on by adding the period (4+0.54=4.54), but i couldnt get the one in between.

- January 30th 2008, 06:05 PMJOhkonutPeriodic Function- find the first three positive values of x...
f(x)=4sin[(pi/2)x]+2

Find the first three positive values of x for f(x)=5.

I was able to get one of the answers(0.54) and another on by adding the period (4+0.54=4.54), but i couldnt get the one in between. - January 30th 2008, 11:51 PMCaptainBlack
- February 2nd 2008, 09:36 AMinvite_moon
f(x)=4sin(pi/2*x)+2=5

so we have

sin(pi/2*x)=3/4

so pi*x/2=arcsin(3/4) --> x=2arcsin(3/4)/pi

and T=2pi/(pi/2)=4 so T/2=2

we have

x1=2arcsin(3/4)/pi

x2=T/2-x1=2-2arcsin(3/4)/pi

so we have

x(2k+1)=2arcsin(3/4)/pi +kT=2arcsin(3/4)/pi+4k k=0.1.2.3.4...

x(2k)=2-2arcsin(3/4)/pi+kT=2-2arcsin(3/4)/pi+4k k=1.2.3.4...