how to show that $\csc x - \csc\left(\frac{\pi}{3} + x \right) + \csc\left(\frac{\pi}{3} - x\right) = 3 \csc 3x$? 
How to show that $\csc x - \csc\left(\frac{\pi}{3} + x \right) + \csc\left(\frac{\pi}{3} - x\right) = 3 \csc 3x$?

My attempt:
\begin{align}
LHS &= \csc x - \csc\left(\frac{\pi}{3} + x\right) + \csc\left(\frac{\pi}{3} - x\right) \\
&= \frac{1}{\sin x} - \frac{1}{\sin\left(\frac{\pi}{3} + x\right)} + \frac{1}{\sin\left(\frac{\pi}{3} -x\right)} \\
&= \frac{\sin x \sin\left(\frac{\pi}{3} + x\right) + \sin\left(\frac{\pi}{3} + x\right) \sin\left(\frac{\pi}{3} - x\right) - \sin\left(\frac{\pi}{3} - x\right) \sin x }{\sin x \sin\left(\frac{\pi}{3} + x\right) \sin\left(\frac{\pi}{3} - x\right)} \\
&=\frac{4}{\sin 3x}\left(\sin x \sin\left(\frac{\pi}{3} + x\right) + \sin\left(\frac{\pi}{3} + x\right) \sin\left(\frac{\pi}{3} - x\right) - \sin\left(\frac{\pi}{3} - x\right) \sin x\right)  \\
&=\frac{4}{\sin 3x}\left(\sin x \sin\left(\frac{\pi}{3} + x\right) - \sin\left(\frac{\pi}{3} - x\right) \left(\sin x - \sin\left(\frac{\pi}{3} + x\right)\right)\right)  \\
&=\frac{4}{\sin 3x}\left(\sin x \sin\left(\frac{\pi}{3} + x\right) - \sin\left(\frac{\pi}{3} - x\right) \left(2\sin\frac{-\pi}{6}\cos\left(x + \frac{\pi}{6}\right)\right)\right)  \\
&=\frac{4}{\sin 3x}\left(\sin x \sin\left(\frac{\pi}{3} + x\right) + \sin\left(\frac{\pi}{3} - x\right) \cos\left(x + \frac{\pi}{6}\right)\right)  \\
\end{align}
How should I proceed? Or did I make some mistakes somewhere? Thanks in advance.
 A: We have $$\sin x \sin\left(\frac{\pi}{3} + x\right) + \sin\left(\frac{\pi}{3} - x\right) \cos\left(x + \frac{\pi}{6}\right)  \\$$
$$=\sin(x)[\sin(\frac{\pi}{3})\cos(x)+\cos(\frac{\pi}{3})\sin(x)]$$
$$+(\sin(\frac{\pi}{3})\cos(x)-\cos(\frac{\pi}{3})\sin(x))(\cos(\frac{\pi}{6})\cos(x)-\sin(\frac{\pi}{6})\sin(x))$$
$$=\sin(x)[\frac{\sqrt{3}}{2}\cos(x)+\frac{1}{2}\sin(x)]$$
$$+(\frac{\sqrt{3}}{2}\cos(x)-\frac{1}{2}\sin(x))(\frac{\sqrt{3}}{2}\cos(x)-\frac{1}{2}\sin(x))$$
$$=\frac{\sqrt{3}}{2}\sin(x)\cos(x)+\frac{1}{2}\sin^2(x)$$
$$+\frac{3}{4}\cos^2(x)-\frac{\sqrt{3}}{2}\cos(x)\sin(x)+\frac{1}{4}\sin^2(x)$$
or $$\frac{3}{4}(\sin^2(x)+\cos^2(x))=\frac{3}{4}.$$
A: $$\frac{1}{\sin{x}}-\frac{1}{\sin\left(\frac{\pi}{3}+x\right)}+\frac{1}{\sin\left(\frac{\pi}{3}-x\right)}=$$
$$=\frac{\sin\left(\frac{\pi}{3}-x\right)\sin\left(\frac{\pi}{3}+x\right)+\sin{x}\left(\sin\left(\frac{\pi}{3}+x\right)-\sin\left(\frac{\pi}{3}-x\right)\right)}{\sin{x}\sin\left(\frac{\pi}{3}+x\right)\sin\left(\frac{\pi}{3}-x\right)}=$$
$$=\frac{\frac{1}{2}\left(\cos2x-\cos\frac{2\pi}{3}\right)+\sin{x}\cdot2\sin{x}\cos\frac{\pi}{3}}{\sin{x}\left(\frac{\sqrt3}{2}\cos{x}+\frac{1}{2}\sin{x}\right)\left(\frac{\sqrt3}{2}\cos{x}-\frac{1}{2}\sin{x}\right)}=$$
$$=\frac{\frac{1}{2}-\sin^2x+\frac{1}{4}+\sin^2x}{\sin{x}\left(\frac{3}{4}\cos^2x-\frac{1}{4}\sin^2x\right)}=\frac{3}{\sin{x}(3(1-\sin^2x)-\sin^2x)}=\frac{3}{\sin3x}.$$
A: Using the identities
$$\sin A \sin B = \frac12 (\cos (A-B) - \cos (A+B))$$
$$\sin A \cos B = \frac12 (\sin (A+B) + \sin (A-B))$$
we have
\begin{align}
&\phantom{=}\sin x \sin\left(\frac{\pi}{3} + x\right) + \sin\left(\frac{\pi}{3} - x\right) \cos\left(x + \frac{\pi}{6}\right)\\&=\frac12\left(\cos \left(-\frac\pi3\right)-\cos\left(2x+\frac\pi3\right)+\sin\frac\pi2+\sin\left(\frac\pi6-2x\right)\right)\\
&=\frac 12\left(\frac12+1-\cos\left(2x+\frac\pi3\right)+\cos\left(\frac\pi2 - \left(\frac\pi6-2x\right)\right)\right)\\
&=\frac34+\frac 12\left(-\cos\left(2x+\frac\pi3\right)+\cos\left(2x+\frac\pi3\right)\right)\\
&=\frac34
\end{align}
A: If $\csc3x=\csc3y\iff \sin3y=\sin3x$
$$3y=3x+(-1)^nn\pi$$ where $n$ is any integer
$y=x+\dfrac{2n\pi}3; n=-1,0,1$
Now as $\sin3y=3\sin y-4\sin^3y,$
the roots of $$4\sin^3y-3\sin y+\sin3x=0$$ are
$\sin\left(x-\dfrac{2\pi}3\right)=\sin\left(-\pi+\left(x+\dfrac\pi3\right)\right)=-\sin\left(x+\dfrac\pi3\right)=x_1,$
$\sin x=x_2,$
$\sin\left(x+\dfrac{2\pi}3\right)=\sin\left(\pi-\left(\dfrac\pi3-x\right)\right)=\left(\dfrac\pi3-x\right)=x_3$
By Vieta's formula,
$$\dfrac1{x_1}+\dfrac1{x_2}++\dfrac1{x_3}=\dfrac{x_1x_2+x_2x_3+x_3x_1}{x_1x_2x_3}=\dfrac{\dfrac{-3}4}{-\dfrac{\sin3x}4}$$
