Please tell me why my method of solving $\sin\left(\frac{\pi}{3}-2\,x\right)-\sin\left(6\,x\right)=2$ is not correct.

Question

$\sin\left(\frac{\pi}{3}-2\,x\right)-\sin\left(6\,x\right)=2$

Since $\left| \sin\left(\frac{\pi}{3}-2\,x\right)\right|\leqslant 1$ and $\left| \sin\left(6x\right)\right|\leqslant 1$, the two sides of the equation can be equal when and only when $\sin\left(6\,x\right)=-1$ and $\sin\left(\frac{\pi}{3}-2\,x\right)=1$ :

$\left\{\begin{matrix}\sin\left(\frac{\pi}{3}-2\,x\right)=1\\\sin\left(6\,x\right)=-1\end{matrix}\right.$

$\left\{\begin{matrix}\frac{\pi}{3}-2\,x=\frac{\pi}{2}+2\,\pi\,n\\6\,x=-\frac{\pi}{2}+2\,\pi\,k\end{matrix}\right.$

$\left\{\begin{matrix}x=-\frac{\pi}{12}-\pi\,n\\x=-\frac{\pi}{12}+\frac{\pi\,k}{3}\end{matrix}\right.$

And then I equate the found values:

$-\frac{\pi}{12}-\pi\,n=-\frac{\pi}{12}+\frac{\pi\,k}{3}$

Getting $n=-\frac{k}{3}$

And then I just substitute $n=-\frac{k}{3}$ into $x=-\frac{\pi}{12}-\pi\,n = -\frac{\pi}{12}+\frac{\pi\,k}{3}$.

And this solution is wrong. I am very curious why, because I solved $\sin\left(3\,x\right)\,\cos\left(4\,x\right)=1$ using this method and the solution was correct. But it doesn't work for this equation.

Answer 1

It's simple. Because $n$ is an integer, you must ensure that $k$ is a multiple of $3$, and you can't arbitrarily value $k$.

Answer 2

We can rewrite system as follows $$\begin{cases} x=-\frac{\pi}{12}+\pi n, n\in \mathbb{Z},\\ x=-\frac{\pi}{12}+\frac{\pi k}{3}, k\in \mathbb{Z}. \end{cases}$$ You need to intersect two series of solutions, but the first series of solutions is contained in the second, because the first series of solutions can be obtained from the second one by substituting instead of $k$ integers of the form $k=3m$, $m\in\mathbb{Z}$. So, the intersection will give us the first series of solutions.

You can also verify this if you draw a trigonometric circle and mark on it the solutions from the first series and the solutions of the second series, which belongs $[0,2\pi)$.