Solve for the positive values of $x$ in $\ln(e^{\frac{1}{2}}+\sin x) = \frac{1}{2}$

I tried:

$$\ln(e^{\frac{1}{2}}+\sin x) = \frac{1}{2} \Leftrightarrow \\ e^{\frac{1}{2}}+\sin x = e^{\frac{1}{2}} \Leftrightarrow \\ 0 = \sin x \Leftrightarrow \\ x = \arcsin(0) +2k\pi \lor x = \pi-\arcsin(0)+2k\pi \Leftrightarrow \\ x = 2k\pi \lor x = \pi+2k\pi$$ $$k \in \mathbb{Z}$$

However, my book says the solution is:

$$x = k\pi$$

Did I do something wrong or is my book's solution just the simplyfied version of mine?

• you solution is the same that the book, you have divided $x$ in your solution in the set of odd multiples of $\pi$ and the even multiples of $\pi$. Together they are the multiples of $\pi$. – Masacroso Apr 27 '17 at 16:54
• Take a look at the graph of $y=\sin x$. You will see that $\sin x=0$ at every integer multiple of $\pi$. – John Wayland Bales Apr 27 '17 at 16:56

If, \begin{align*} \sin\theta&=x\\ \Rightarrow \sin^{-1}x&=n\pi+(-1)^n\theta\hspace{25pt}\text{ where, }n\in\mathbb Z \end{align*} This is the General Value of $\sin^{-1}x$, and the Principal Value of $\sin^{-1}x$ is $\theta$ only.
The values of $x$ for which $\sin x=0$ are $0,\,\pm\pi,\,\pm2\pi,\, \pm 3\pi, \ldots,$ i.e. $k\pi$ for $k\in \mathbb Z.$