If $2^9.{\sin }^6\theta .{\cos }^4\theta =2\cos 10\theta -2\cos 8\theta -3\cos 6\theta +8.\cos 4\theta +2\cos 2\theta -k$, then $k=$

$x=\cos \theta +i\sin \theta \Rightarrow \frac{1}{x}=\cos \theta -i\sin \theta$

$x+\frac{1}{x}=2\cos \theta$ and $x-\frac{1}{x}=2i\sin \theta$

${\left(2i\sin \theta \right)}^6.{\left(2\cos \theta \right)}^4={(x+\frac{1}{x})}^4.{(x-\frac{1}{x})}^6$

$\Rightarrow -2^6.{\sin }^6\theta {.2}^4.{\cos }^4\theta ={(x^2-\frac{1}{x})}^4{(x-\frac{1}{x})}^2$

$={[x^4+\frac{1}{x^4}-2]}^2[x^2+\frac{1}{x^2}-2]$

$=(x^8+\frac{1}{x^8}+4+2-\frac{4}{x^4}-4.x^4)(x^2+\frac{1}{x^2}-2)$

$=x^{10}+x^6-2x^8+\frac{1}{x^6}+\frac{1}{x^{10}}-\frac{2}{x^8}+6x^2+\frac{6}{x^2}-12$

$-\frac{4}{x^2}-\frac{4}{x^6}+\frac{8}{x^4}-4x^6-4x^2+8x^4$

$=[x^{10}+\frac{1}{x^{10}}]-2[x^8+\frac{1}{x^8}]-3[x^6+\frac{1}{x^6}]+8(x^4+\frac{1}{x^4})+2(x^2+\frac{1}{x^2})-12$

$=2.\cos 10\theta -2.\cos 8\theta -3.\cos 6\theta +8.\cos 4\theta +2.\cos 2\theta -12$