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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?

$$x = \frac \pi {32} + \frac {\pi}{2}k$$

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  • $\begingroup$ $\sin(\frac{3}{8}\pi=3x)$ is a typo, isn't it? $\endgroup$ Mar 5 '16 at 15:40
  • $\begingroup$ Sorry about that. Fixed! $\endgroup$
    – Cesare
    Mar 5 '16 at 15:41
  • $\begingroup$ We need $1+16k'=-6+32k$ for the second case : is that possible? $\endgroup$ Mar 5 '16 at 15:44
  • $\begingroup$ @labbhattacharjee I don't understand where the equation you just wrote comes from. $\endgroup$
    – Cesare
    Mar 5 '16 at 15:46
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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.

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  • $\begingroup$ Excellent answer. Thanks! :) $\endgroup$
    – Cesare
    Mar 5 '16 at 15:57
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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 $16n+32m=-7$.

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  • $\begingroup$ I see where you're coming from. Thanks for your answer! $\endgroup$
    – Cesare
    Mar 5 '16 at 15:58

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