Ex $P(t)=2.95\cdot \sin[(\frac{\pi}{6})(x+C)]+3.15$
So to find the phase shift I would look for the lesser of the two infinite pairs of input points because these will give me points that begin one of the cycles of a positive sine function. So to do this I make $P(t)$ equal to the mid line or $P(t)= 2.95$
As I solve I'll get stuck with having to solve for $x+C$:
$2.95=2.95\cdot \sin[(\frac{\pi}{6})(x+C)]+3.15$
$-0.2=2.95\cdot \sin[(\frac{\pi}{6})(x+C)]$
$(\frac{-0.2}{2.95})= \sin[(\frac{\pi}{6})(x+C)]$
$\sin^{-1}(\frac{-0.2}{2.95})=(\frac{\pi}{6})(x+C)$
$(\frac{6}{\pi})[\sin^{-1}(\frac{-0.2}{2.95})]=x+C$
$\frac{6}{\pi}\sin^{-1}\frac{-0.2}{2.95} = -.12958$
For $C = 0$, when $x = -.12958, P(t) = 2.95$
A $C$ value will shift the $2.95$ point to the left by +C and to the right by -C. So if you wanted to bring the $2.95$ point onto the $y$ axis $(x = 0)$, make $C = -.12958$
That is $P(t) = 2.95\cdot \sin(\frac{\pi}{6}(x - .12958))+3.15$
When $x = 0, P(t) = 2.95$