Trouble with this one step involving arcsin So I'm solving a very basic trigonometric equation and after substituting the function for u and solving the quadratic equation, I ended up with $\sin\left(2x+\frac{\pi}{3}\right)=\frac{1}{4}$.
I need to take the arcsin of both sides if I'm not mistaken, but after inputting the original equation into a graph, it gives me a different solution than $2x+\frac{\pi}{3}=\arcsin\left(\frac{1}{4}\right)$
So 100% I'm missing a rule regarding cancelling out a function. I would appreciate an elaboration :)
 A: $\arcsin\frac14$ is a principal value, and you can add any integer multiple of $2\pi$ to it – and replace $y$ by $\pi-y$. Hence it should be
$$\arcsin\frac14+2n\pi,n\in\mathbb Z$$
$$\pi-\arcsin\frac14+2n\pi,n\in\mathbb Z$$
A: The number $\arcsin\frac14$ is the only number in $\left[-\frac\pi2,\frac\pi2\right]$ whose sine is $\frac14$. And $\pi-\arcsin\frac14$ is the only number in $\left[\frac\pi2,\frac{3\pi}2\right]$ whose sine is $\frac14$. The remaining numbers with that property are those of the form $2k\pi+\arcsin\frac14$ or of the form $2k\pi+\pi-\arcsin\frac14$, with $k\in\Bbb Z$.
A: If we have $\sin(\alpha) = \frac14$
We have $\alpha=2k\pi + \arcsin\frac14$ or $\alpha=(2k-1)\pi-\arcsin\frac14, k\in \mathbb{Z}$.
Hence $2x + \frac{\pi}3=2k\pi + \arcsin\frac14$ or $2x+\frac{\pi}3=(2k-1)\pi-\arcsin\frac14, k \in \mathbb{Z}$.
A: 
For $\alpha \in (-1,1)$ the equation $\sin(x) = \alpha$ has solutions
$$ x = \arcsin(\alpha) + 2 k \pi \qquad \text{and}\qquad   x = \pi - \arcsin(\alpha) + 2 k \pi $$

Hence, in your case, the solutions are
$$
2x + \frac{\pi}{3}= \arcsin\left(\frac{1}{4}\right)+ 2 k \pi \implies x = -\frac{\pi}{6} +\frac{1}{2}\arcsin\left(\frac{1}{4}\right) + k\pi\\
$$
and
$$
2x + \frac{\pi}{3} = \pi - \arcsin\left(\frac{1}{4}\right) + 2 k \pi \implies x = -\frac{2\pi}{3} -\frac{1}{2}\arcsin\left(\frac{1}{4}\right) + k\pi
$$
where $\arcsin\left(\frac{1}{4}\right) \approx 0.2526$.
