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.
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.
Tell us why it is notQuote:
Originally Posted by JOhkonut
, where
is the first solution greater than zero.
RonL
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...