Solving $\sin5x \cos3x = \sin6x \cos2x$ two ways gives different solutions. Which approach is correct? The question is:

$$\sin5x \cos3x = \sin6x \cos2x$$

I had two approaches:

*

*$$\sin5x \cos3x = \sin6x \cos2x \\
   2\sin5x \cos3x = 2\sin6x \cos2x \\
   \sin8x+\sin2x=\sin8x+\sin4x \\
   \sin2x=\sin4x \\ \sin4x-\sin2x=0 \\
 2\sin x\cos3x=0  \\
\implies \sin x=0 \\ x=n\pi
\implies  \cos3x=0 \\ 3x={(2n+1)\pi}/2 \\
x={(2n+1)\pi}/6$$
These were the solutions in the first approach.


*$$\sin5x \cos3x = \sin6x \cos2x \\
   2\sin5x \cos3x=2\sin6x\cos2x \\
   \sin8x+\sin2x=\sin8x+\sin4x \\
   \sin2x=\sin4x\\ \sin4x-\sin2x=0 \\
   2\sin2x.\cos2x-\sin2x=0 \\
\sin2x(2\cos2x-1)=0 \\
\implies \sin2x=0 \\
x=n(π/2)\\
\implies 2\cos2x-1=0 \\
\cos2x=1/2 \\
2x=2n\pi\pm(\pi/3) \\
x=n\pi\pm(\pi/6)$$
These were my solutions in the second approach. This was also the approach given in the textbook for this question.  
My question is:

Why do these solutions not match in the two cases? Have I made any mistake in the first approach?

 A: The two solutions are equivalent - they both include all numbers which are congruent to $0$, $\pi/6$, $\pi/2$ or $5\pi/6$ mod $\pi$. It is just that in your first solution you have divided into the first case and a combination of all others, whereas your other solution groups alternate cases together.
A: Your solutions are equivalent, indeed for $x\in [0,2\pi)$

*

*$x=n\pi \land x={(2n+1)\pi}/6 \implies x\in\{0,\frac \pi 6,\frac \pi 2,\frac 2 3 \pi,\pi,\frac 4 3 \pi,\frac 32\pi,\frac {11}6\pi\}$


*$x=n\frac \pi 2 \land x=n\pi\pm(\pi/6) \implies x\in\{0,\frac \pi 6,\frac \pi 2,\frac 2 3 \pi,\pi,\frac 4 3 \pi,\frac 32\pi,\frac {11}6\pi\}$
As an alternative by product to sum identities we have that
$$\sin(5x)\cos(3x)=\frac12\sin(8x)+\frac12\sin(2x)$$
$$\sin(6x)\cos(2x)=\frac12\sin(8x)+\frac12\sin(4x)$$
then
$$\sin(5x)\cos(3x)=\sin(6x)\cos(2x) \iff \sin(2x)=\sin(4x)$$
and use that
$$\sin A=\sin B \iff A=B+2k\pi \lor A=\pi - B+2k\pi$$
that is
$$x=k\pi \lor x=\frac \pi 6+k\frac \pi 3$$
which are equivalent to both your solutions.
A: Let $z=e^{i\theta}$.
$$(z^5-z^{-5})(z^3+z^{-3})=(z^6-z^{-6})(z^2+z^{-2})$$
expands as
$$z^2-z^{-2}=z^4-z^{-4}$$ or $$\sin2\theta=\sin 4\theta=2\sin2\theta\cos2\theta.$$
Hence you have the roots of $\sin2\theta$, $\dfrac{k\pi}2$, and the solutions of $\cos2\theta=\dfrac12$, $\pm\dfrac\pi6+k\pi$.
In terms of multiples of $\dfrac\pi6$,
$$0,3,5,6,7,9,11,12,13,15,17,18,19,21,23,24,25,\cdots$$
A: Case$\#1:$ $$\dfrac{m\pi}2=\dfrac{(2n+1)\pi}6$$
$$\iff2n+1=3m\iff2(n-1)=3(m+1)\iff3|(n-1)\iff n\equiv1\pmod3$$
Case$\#2:$
$$\dfrac{(2n+1)\pi}6=m\pi\pm\dfrac\pi6=\dfrac{\pi(6m\pm1)}6\iff2n+1=6m\pm1$$
For $'+',$ $$2n+1=6m+1\iff n=3m, n\equiv0\pmod3$$
For $'-',$ $$2n+1=6m-1\iff n=3m-1, n\equiv-1\pmod3$$
So, all three possible values of $n$ are covered.
