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To solve this I did $x = \arcsin(\frac{1}{2}) = \frac{\pi}{6}$

However, my book states that the solution is $$x = \frac{\pi}{6}+k2\pi \lor x = \frac{5\pi}{6}+k2\pi, k \in \mathbb{Z}$$

What did I do wrong?

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  • $\begingroup$ well, arcsin always gives you only one solution, but of course sin is periodic so there's a lot more solutions than the one the arcsin gives you.. as a matter of fact, the arcsin always gives you the unique solution between $-\pi /2$ and $\pi/2$ $\endgroup$ – Sebastian Schulz Apr 6 '17 at 0:18
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The arcsine function $$\arcsin x: [-1, 1] \to \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$$ is defined by $$\arcsin x = y \iff \sin y = x, y \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$$ By finding $$x = \arcsin\left(\frac{1}{2}\right) = \frac{\pi}{6}$$ you found the unique value of $x$ in the interval $[-\pi/2, \pi/2]$ such that $\sin x = 1/2$. However, the solution set of the equation $$\sin x = \frac{1}{2}$$ is the set of all angles that have sine $1/2$.

An angle in standard position (vertex at the origin, initial side on the positive $x$-axis) has sine $1/2$ if the terminal side of the angle intersects the unit circle at a point that has $y$-coordinate $1/2$. You have shown that one such angle is $\pi/6$. To find the others, consider the following diagram.

symmetry_diagram_for_sine_and_cosine

Two angles have the same sine if the $y$-coordinates of the terminal side of the angle are equal. Thus, by symmetry, $\sin(\pi - \theta) = \sin\theta$. Also, coterminal angles have the same sine since they intersect the unit circle at the same point. Hence, $$\sin\theta = \sin\varphi$$ if $$\varphi = \theta + 2k\pi, k \in \mathbb{Z}$$ or $$\varphi = \pi - \theta + 2k\pi, k \in \mathbb{Z}$$ We wish to solve the equation $$\sin x = \frac{1}{2}$$ Since a particular solution is $$x = \arcsin\left(\frac{1}{2}\right) = \frac{\pi}{6}$$ we need to find all values of $x$ that satisfy the equation $$\sin x = \sin\left(\frac{\pi}{6}\right)$$ Using the formulas above for the solution of the equation $\sin\theta = \sin\varphi$, we obtain $$x = \frac{\pi}{6} + 2k\pi, k \in \mathbb{Z}$$ or \begin{align*} x & = \pi - \frac{\pi}{6} + 2k\pi, k \in \mathbb{Z}\\ & = \frac{5\pi}{6} + 2k\pi, k \in \mathbb{Z} \end{align*}

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So you are right that a solution to $sin(x)=\frac{1}{2}$ is $x=\frac{\pi}{6}$, however, note that this is far from the only solution. If we look at the geometric definition of the $sin(x)$ function, we see that there are many values of $x$ for which the y-value of the unit circle are 1. The first is $x=\frac{\pi}{6}$, however, that is only the coordinate in the first quadrant. There is also a coordinate $(\frac{-\sqrt3}{2},\frac{1}{2})$ which will come when $x=\frac{5\pi}{6}$. Now also note that a rotation of $2\pi$ will get you back to the same value, and so the solutions are not only $x=\frac{\pi}{6}$ and $x=\frac{5\pi}{6}$, but $x=\frac{\pi}{6}+2\pi k$ and $x=\frac{5\pi}{6}+2\pi k, k \in \mathbb{Z}$.

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  • $\begingroup$ So, for every equation of this kind there are always two solutions? $\endgroup$ – Mark Read Apr 6 '17 at 0:23
  • $\begingroup$ For essentially all equations with $sin(x)=a$ for some value a, yes. The only exceptions are the maximum and minimum values of $f(x)=sin(x)$, 1 and -1. Have a look at this. You can see that within any one period of the sin(x) function there will be 2 points of for a non-zero value between 0 and 1 (as shown with $y=\frac{1}{2}$). The only exceptions are y=-1, y=1 (as they are extremities) and y=0 (though I'm not completely sure if that counts as it passes through three times if the domain is 0 to 2pi inclusive). $\endgroup$ – Cameron Eggins Apr 6 '17 at 0:38
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Hint:

Use the properties: $\;\sin(\pi-x)=\sin x$ and $\sin x$ has period $2\pi$. There results the equation $\sin x=\sin \theta$ has as solutions: $$\begin{cases} x\equiv\theta\\x\equiv\pi-\theta \end{cases}\mod2\pi$$

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