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?

  • 2
    $\begingroup$ 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 $\pi$. $\endgroup$ – Masacroso Apr 27 '17 at 16:54
  • 1
    $\begingroup$ Take a look at the graph of $y=\sin x$. You will see that $\sin x=0$ at every integer multiple of $\pi$. $\endgroup$ – 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.

  • $\begingroup$ Interesting and simpler than what I have been using, thanks. $\endgroup$ – Mark Read Apr 27 '17 at 17:00

The values of $x$ for which $\sin x=0$ are $0,\,\pm\pi,\,\pm2\pi,\, \pm 3\pi, \ldots,$ i.e. $k\pi$ for $k\in \mathbb Z.$

Other than that, your solution looks good.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.