Finding general solution of trigonometric equation $\sin(x+\pi/6)=\sin(x+\pi)$

Question

Consider the trigonometric equation $\sin(x+\pi/6)=\sin(x+\pi)$

$\sin(x+\pi)$ can be reduced to $-\sin(x)$ which can further be written as $\sin(-x)$. Now the formula for general solution can be applied and all the general solutions can be obtained.

However,this can be approached yet another way but the result do not fits as the solution. What if we don't reduce $\sin(x+\pi)$ and apply the formula $\sin (x)=\sin(y) \to x=n\pi +(-1)^n y$ on the initial equation. I don't see any mathematical fallacy in the above argument yet the answers don't match.

Thanks in advance.

Answer 1

hint

If you apply the formula, you will get as solutions

$$x+\frac{\pi}{6}=n\pi+(-1)^n(x+\pi)$$

the solution exist only for odd $n$.

Answer 2

Your argument is valid. Nevertheless in case of your attempt for $\sin(x)=\sin(y)$ you have to consider two cases, the one for odd $n$ and the one for even $n$. Therefore you will get

$$\sin(x+\pi/6)=\sin(x+\pi)\Leftrightarrow x+\pi/6=n\pi+(-1)^n (x+\pi)$$

For $n=2k, (k\in\mathbb{Z})$ this leads to a contradiction

$$x+\pi/6=2k\pi+(x+\pi) \Leftrightarrow \pi/6 \ne (2k+1)\pi$$

For $n=2k+1, (k\in\mathbb{Z})$ you will get

$$x+\pi/6=(2k+1)\pi-(x+\pi) \Leftrightarrow x = -\pi/12+k\pi$$

which is identically to the solution you would get by going with your first attempt.

Answer 3

Leave the $(-1)^ny$ alone. The equality $\sin\alpha=\sin\beta$ is true if and only if $$ \alpha=\beta+2k\pi \qquad\text{or}\qquad \alpha=\pi-\beta+2k\pi $$ (as usual, $k$ denotes an arbitrary integer).

In your case, the first possibility becomes $$ x+\frac{\pi}{6}=x+\pi+2k\pi $$ which is clearly impossible.

Thus you remain with $$ x+\frac{\pi}{6}=\pi-(x+\pi)+2k\pi $$ that is $$ 2x=-\frac{\pi}{6}+2k\pi $$ and so $$ x=-\frac{\pi}{12}+k\pi $$

If you transform the equation into $\sin(x+\pi/6)=\sin(-x)$, the possibilities would be either $$ x+\frac{\pi}{6}=-x+2k\pi $$ or $$ x+\frac{\pi}{6}=\pi-(-x)+2k\pi $$ The former has the solutions $$ x=-\frac{\pi}{12}+k\pi $$ whereas the latter has no solution. Nothing changes, as you see.