a minimum value of a sinusoidal function is at (pi/4,3). The nearest maximum value to the right of this point is at (7pi/12,7). Determine an equation of this function
So 7pi/12 - pi/4 gives half the period.
And 7-3 gives twice the amplitude, where amplitude is distance from center position to either maximum or minimum.
So you'll want to transform $\displaystyle \sin x$ as $\displaystyle a\sin(bx)+c$ to fit those requirements (or you could use cosine).
let
f(x)=Asin(ax+b)+c
at (pi/4,3) we have minimum value for this sin(ax+b) should be minimum that is equal to -1
3=-A+c............................................... .....1
at (7pi/12,7) we have maximum value for that sin(ax+b) must be maximum that is equal to 1 therfor
7=A+c............................................. ........2
solving equation 1 and 2 we get
A=2
c=5
now
f(x)=2sin(ax+b)+5
we know two points that are (pi/4,3) and (7pi/12,5) using them we get
1) 3=2sin(api/4+b)+5 or
sin(api/4+b)=-1
api/4+b=-pi/2.......................................3
2) similarly for (7pi/12,7) we get
7api/12+b=pi/2......................................4
solving equation 3 and 4 we get
a=3 and b=-5pi/4
therfor
f(x)=2sin(3x-5pi/4)+5