For the question $\sin6x=\sin4x-\sin2x$, I applied $\sin C-\sin D$ formula on RHS, and got three general solutions of x: $(2l+1)\pi/6$, $(2m+1)\pi/4$, $n\pi$. But my book and so many websites have solved it by clubbing $\sin6x$ and $\sin2x$ applying $\sin C+\sin D$. They got two general solutions of x: $m\pi/4$, $(6n\pm 1)\pi/6$. For the interval $0$ to $2\pi$, I have verified that both the methods are giving the same values. However, I have a couple of questions on this:
- Why are we getting different formats of answers with different methods?
- Is it preferable to use $\sin C+\sin D$ over $\sin C-\sin D$, or to use bigger angle ($\sin6x$) first while solving?
- How to reach the two answer format from my three answer format?
- Can we club all the values to get just one general solution?
Let's compare $$\dfrac{(2a+1)\pi}6,\dfrac{(2b+1)\pi}4,e\pi$$ with $$\dfrac{c\pi}4,\dfrac{(6d\pm1)\pi}6$$
Case $\#1: \dfrac{(2a+1)\pi}6=\dfrac{c\pi}4\iff2(2a+1)=3c$
$\iff3c-4a=2(4-3)\iff3(c+2)=4(2+a)$
$\implies4\mid(c+2)\iff c\equiv-2\pmod4\equiv2$
and $3|(2+a)\iff a\equiv-2\pmod3\equiv1$
Case $\#2: \dfrac{(2b+1)\pi}4=\dfrac{c\pi}4\iff c=2b+1$
Case $\#3:e\pi=\dfrac{c\pi}4\iff c=4e$
Case $\#4: \dfrac{(2a+1)\pi}6=\dfrac{(6d\pm1)\pi}6\iff2a+1=6d\pm1$
$\implies a=3d$ or $a=3d-1$
Case $\#5: \dfrac{(2b+1)\pi}4=\dfrac{(6d\pm1)\pi}6\iff3(2b+1)=2(6d\pm1)$
$6b+3=12d+2\iff1=6(2d-b)$ which is untenable
or $6b+3=12d-2\iff5=6(2d-b)$ which is also untenable
Case $\#6:e\pi=\dfrac{(6d\pm1)\pi}6\iff6e=6d\pm1$ which is also untenable