Solutions of the trigonometric equation $\sin(x+ \frac \pi 4) = \sin(\frac 3 8 \pi-3x)$ The following equation
$$\sin(x+ \frac \pi 4) = \sin(\frac 3 8 \pi-3x)$$
has these solutions:
$$x = \frac {\pi} {32} + k \frac \pi 2 \space \vee x = -\frac 3 {16} \pi + k\pi$$
Is it safe to say the equation has this solution below? Did I group the solutions above correctly?
$$x = \frac \pi {32} + \frac {\pi}{2}k$$
 A: The equality $\sin\alpha=\sin\beta$ is equivalent to the condition
$$
\alpha=\beta+2k\pi\quad\text{or}\quad \alpha=\pi-\beta+2k\pi
$$
(where $k$ is an integer). So in your case you have
$$
x+\frac{\pi}{4}=\frac{3\pi}{8}-3x+2k\pi
$$
which gives
$$
\boxed{x=\frac{\pi}{32}+k\frac{\pi}{2}}
$$
or the other branch
$$
x+\frac{\pi}{4}=\pi-\frac{3\pi}{8}+3x+2k\pi
$$
that yields
$$
x=-\frac{3\pi}{16}-k\pi
$$
that you can also write as
$$
\boxed{x=-\frac{3\pi}{16}+k\pi}
$$
since $k$ can be an arbitrary integer.
Can you group into just the first solution set? You'd need to write
$$
-\frac{3\pi}{16}=\frac{\pi}{32}+k\frac{\pi}{2}
$$
for some integer $k$, but the equation gives
$$
-6=1+16k
$$
that has no integer solution.
A: If you can represent $x=-\frac{3}{16}\pi+k\pi$ as $\frac{\pi}{32}+\frac{k'\pi}{2}$ for some $k'\in\mathbb{Z}$, then you are right. Then, how about $x=\frac{13}{16}\pi$? Set equation $\frac{\pi}{32}+\frac{k\pi}{2}=\frac{13}{16}\pi$, then we get $k=\frac{25}{32}$, which is not an integer. Thus you are not right.
In fact, there are no integers $n,m$ such that $\pi/32 + n\pi/2=-3\pi/16+m\pi$, since there is no integer root of $16n+32m=-7$.
