Below is my wrong solution:
By the formula $\arccos x + \arcsin x = \pi / 2$, for $ x \in [-1, 1]$, I have
$$ y = \frac{\pi}{2} - \arcsin(\sin x) \text{ (because $\sin x \in [-1, 1]$)} \\ \implies y = \frac{\pi}{2} - x, \ \text{for $x \in R$} $$
But the correct solution in my book is
May anyone tell me why I am wrong and how to solve the question?
The problem is that $\arcsin(x)$ is not equal to $x$ for all $x \in \mathbb R$. This is because $\arcsin$ is defined as $$\arcsin : [-1,1] \to [-\pi/2,\pi/2]$$ You can see from here that $\arcsin$ can only have values from $[-\pi/2, \pi/2]$. But, the expression $y=\frac{\pi}{2}-x$ can have any real value.
What you need to do is to somehow restrict $x$ to the interval $[-\pi/2,\pi/2]$. You do this by adding a constant, but that constant depends on the interval $I\subset \mathbb R$ that you are analyzing.
One thing I like to do with functions like these is to first differentiate and then integrate the function.
Namely, we have $$(\arccos(\sin x))' = -\frac{\cos x}{\sqrt{1-\sin^2x}} = -\frac{\cos x}{|\cos x|} = -\text{sgn}(\cos x)$$ for $\cos x \neq 0$. Then, by integrating we get $$y(x)=\arccos(\sin(x)) = -\int\text{sgn}(\cos x)\ dx = -x\cdot\text{sgn}(\cos x) +C$$ Because $\arccos'(\sin x)$ is not defined for $\cos x = 0$ i.e. $x=(k+\frac{1}{2})\pi$, $k\in \mathbb Z$, this integral is not defined in those points. Keep in mind that this constant depends on the interval $[(k+\frac{1}{2})\pi,(k+\frac{3}{2})\pi]$ that you are considering. To find the constant for each interval, use the fact that $y(x)$ is continuous and find some characteristic points that you can easily evaluate the function at. In your specific problem, determine the constant such that $$-x_k\cdot\text{sgn}(\cos x_k) +C=0$$ where $$x_k = \frac{4k+1}{2}\pi$$