Trigonometry conversion rules, why this way? We have domain $\,[0, 2\pi]\,$ and the following functions are given:
$$f(x)=\cos(2x) \text{ and } g(x)=\sin(x-\pi/3)$$Solve exactly: $\,f(x)=g(x)$
Why does one solve:
(right way)
$$\cos(2x)=\sin(x-\pi/3)\\\cos(2x)=\cos(\pi/2-(x-\pi/3))\\ \text{etc.}\ldots
$$
and not (which gives the wrong answer $\rightarrow a - k\,\, \times\,\, 2\pi \cdots$):
(wrong way)
$$\cos(2x)=\sin(x-\pi/3)\sin(\pi/2-(2x))=\sin(x-\pi/3)\\ \text{etc.}\ldots$$

Right way fully worked out:
$$\cos(2x) = \sin(x -\pi/3)\\
\cos(2x) = \cos(\pi/2 - (x -\pi/3))\\
\cos(2x) = \cos(\pi/2 - x + \pi/3)\\
\cos(2x) = \cos(5\pi/6 - x)\\
2x = (5\pi/6 - x) + k\,\, \times \,\, 2\pi\\
3x = 5\pi / 6 + k \,\, \times \,\, 2\pi\\
x = 5\pi /18 + k \,\, \times \,\, 2\pi\\ 
or:\\
2x = - (5\pi/6 - x) + k\,\, \times \,\, 2\pi\\
x = -5\pi / 6 + k \,\, \times \,\, 2\pi\\
$$

Wrong way fully worked out:
$$\cos(2x) = \sin(x-\pi/3)\\
sin(\pi/2 - x)=\sin(x-\pi/3)\\
\pi/2-2x=(x-\pi/3)+k\,\, \times \,\, 2\pi\\
-3x = (-\pi/3 - \pi/2) + k\,\, \times \,\, 2\pi\\
-3x = -5\pi/6+ k\,\, \times \,\, 2\pi\\
x = 5\pi/18 -k \,\, \times \,\, 2\pi\\
or:\\
\pi/2-2x=\pi-(x-\pi/3) + k \,\, \times \,\, 2\pi\\
\pi/2 - x = \pi - x + \pi/3 + k \,\, \times \,\, 2\pi\\ 
x= 4\pi/3-\pi/2 + k \,\, \times \,\, 2\pi\,\,\,\\\ \,\,\, =5\pi/6 - k \,\, \times \,\,2\pi
$$

On closer inspection of the worked out examples above you can clearly see that final answers of the wrong way are.. wrong. This because the answers we get aren't within the set domain.
Help is highly appreciated.
-Bowser
 A: You can choose to transform the sine into cosine or conversely.
First method
\begin{gather}
\cos2x=\cos\left(\frac{\pi}{2}-\left(x-\frac{\pi}{3}\right)\right)
\\[6px]
\cos2x=\cos\left(\frac{5\pi}{6}-x\right)
\\[6px]
2x=\frac{5\pi}{6}-x+2k\pi
\qquad\text{or}\qquad
2x=-\frac{5\pi}{6}+x+2k\pi
\\[6px]
x=\frac{5\pi}{18}+k\frac{2\pi}{3}
\qquad\text{or}\qquad
x=-\frac{5\pi}{6}+2k\pi
\end{gather}
Second method
\begin{gather}
\sin\left(\frac{\pi}{2}-2x\right)=\sin\left(x-\frac{\pi}{3}\right)
\\[6px]
\frac{\pi}{2}-2x=x-\frac{\pi}{3}+2k\pi
\qquad\text{or}\qquad
\frac{\pi}{2}-2x=\pi-x+\frac{\pi}{3}+2k\pi
\\[6px]
x=\frac{5\pi}{18}-k\frac{2\pi}{3}
\qquad\text{or}\qquad
x=-\frac{5\pi}{6}-2k\pi
\end{gather}
What's happening?
Nothing strange. In the solutions above, $k$ denotes an arbitrary integer: for any choice of the integer $k$, you get a solution. So writing $-k$ or $+k$ is irrelevant: the solution for $k=1$ in the first set (on the left side) appears in the solutions of the second set (left side) for $k=-1$.
The two forms give the same solutions.
If you want to find the solutions in the interval $[0,2\pi)$ you can do as follows, for the first method:
$$
0\le\frac{5\pi}{18}+k\frac{2\pi}{3}<2\pi
\iff
0\le5+12k<36
\iff
-5\le12k<31
$$
that gives $k=0,1,2$. Also
$$
0\le-\frac{5\pi}{6}+2k\pi<2\pi
\iff
0\le-5+12k<12
\iff
5\le12k<17
$$
that gives $k=1$. So you have four solutions.
The same procedure in the second method would give $k=0,-1,-2$ in one case and $k=-1$ in the other.
