How to find $K$ in $2\csc 30 ^\circ(\cos 8x+\cos 6x)(\cos 6x + \cos 2x)-1=\sin (Kx)\cdot \csc x$? The problem is as follows:
Which value should have $K$ in the expression from below so it makes an identity?
$2\csc 30 ^\circ(\cos 8x+\cos 6x)(\cos 6x + \cos 2x)-1=\sin (Kx)\cdot \csc x$
The alternatives given in my book are as follows:
$\begin{array}{ll}
1.&11\\
2.&12\\
3.&14\\
4.&15\\
\end{array}$
I'm confused exactly how to get that particular value for $K$.
The only thing which I was able to spot is that: (Using manipulations on the left side of the equation)
$2\csc 30 ^\circ(\cos 8x+\cos 6x)(\cos 6x + \cos 2x)-1=\sin (Kx)\cdot \csc x$
$2\csc 30 ^\circ (2\cos 7x \cos x)(2\cos 4x \cos 2x)-1$
$4 (2\cos 7x \cdot \cos 2x)(2 \cos x \cdot 4x)-1$
$4(\cos 9x + \cos 5x)(\cos 5x+\cos 3x)-1$
$4(\cos 9x \cos 5x +\cos 9x \cos 3x + \cos 5x \cos 5x + \cos 5x \cos 3x)-1$
$2(\cos 14x + \cos 4x + \cos 11x + \cos 6x + 1+\cos 10x + \cos 8x + \cos 2 x) - 1$
However at the point I reached here then I got stuck it seems that this might not be the intended strategy. What could be done here to solve this problem?. Is there any trick or what?.
Could someone help me with an orderly method to solve this without much fuss?.

If it is an identity, it will be true for any small angle $x$.
Using Taylor series, we have
$$\cos(nx)=1-\frac{n^2 }{2}x^2+O\left(x^4\right)$$ This makes the lhs to be
$$4 (1-32 x^2+1-18 x^2)(1 - 18 x^2+1-2 x^2)-1=15-560 x^2+O\left(x^4\right) \tag 1$$
Now, for the rhs
$$\sin(Kx)=K x-\frac{K^3 }{6}x^3+O\left(x^5\right)$$
$$\csc(x)=\frac{1}{x}+\frac{x}{6}+O\left(x^3\right)$$
$$\sin(Kx)\csc(x)=K+\frac{1}{6} \left(K-K^3\right) x^2+O\left(x^3\right)\tag 2$$
Compare $(1)$ and $(2)$. This gives $K=15$ and you can check that, for $K=15$
$$\frac{1}{6} \left(K-K^3\right)=-560$$
