<s>[INST] Find $b$ in $2\sin(\frac{x}{2}+b)$ given the graph Following is the graph of the function. $2\sin \left(\dfrac{x}{2}+b\right)$

I tried solving the equation using the point $\left (\dfrac{\pi}{2},2\right)$:
$$2 = 2\sin\left(\dfrac{x}{2}+b\right) \Leftrightarrow \\
1 = \sin\left(\dfrac{x}{2}+b\right)\Leftrightarrow \\
\frac{x}{2}+b = k\pi + (-1)^k\arcsin(1) \Leftrightarrow \\
b = 2k\pi$$
But my book $b = \dfrac{\pi}{4}$. How do I solve this?
 [/INST] 
How did you get $b=2k\pi$?

Put $x=\dfrac{\pi}{2}$ in the equation.
$$\frac{x}{2}+b = k\pi + (-1)^k\arcsin(1)$$
Since $\arcsin(1)=\dfrac{\pi}{2}$
You'll get $b+\dfrac{\pi}{4}=k\pi+
(-1)^k\dfrac{\pi}{2}$
Since we have added $b$ in the argument of sine, that is $2b$ in the $x$,  the graph has been shifted backwards by $2b$ units.
Now put $k=1$ , you'll get $b=\dfrac{\pi}{4}$You can try $k=0$ and $k=2$ also, but that all will result in a value of $b$ which isn't in the range $(0,\pi/2)$
(Since any other value of $k$ will lead $b$ to get out of the range $(0,\pi/2)$)
Alternatively since the first negative $x$-intercept of graph is $2b$, and it is clearly visible that it is $\pi/2$, we can say that $$2b=\dfrac{\pi}{2} \implies b=\dfrac{\pi}{4}$$
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