Question 590664
your equation is:
 f(x) = 4cos(3x)+1
you want to  shift it to the right by pi/12.
this means that if x is equal to pi/12, you want to get the cosine of x = 0.
the result of that is that the cosine of 0 is now the cosine of pi/12 which means that value has been shifted to the right by pi/12.
the general formula for a trigonometric function is:
y = a*cos(b(x-c))+d
a is the amplitude.
b is the frequency
c is the horizontal shift
d is the vertical shift of the center line.
your equation has an amplitude of 4 and a vertical shift of 0.
your equation has a frequency of 3 which means the period is equal to 2*pi/3 which makes it equal to (2/3)*pi instead of the 2*pi that it normally is.
based on what you have, the standard form of your equation would be:
f(x) = 4*cos(3*(x-0))+1
a graph of your equation looks like this:
{{{graph(600,600,-7,7,-7,7,4*cos(3x)+1,150*(x-(pi/12)),5)}}}
a vertical line is shown at x = pi/12 to show you where your shift will occur.
since you want to shift the graph to the right, then the value of cosine at x = 0 will then become the value of cosine at x = pi/12.
to do this, your equation has to go from:
f(x) = 4*cos(3*(x-0))+1
to:
f(x) = 4*cos(3*(x-(pi/12)))+1
the graph of that equation looks like this:
{{{graph(600,600,-7,7,-7,7,4*cos(3*(x-(pi/12)))+1,150*(x-(pi/12)),5)}}}