Sufficiently Complex

since true for all x..just sub x=0

$\begin{aligned} 32 \sin^4 x \cos^2 x & = & 32 (\frac{1 - \cos2x}{2})^2 (\frac{1 + \cos2x}{2}) \\ & = & 32(\frac{1 - \cos2x}{2}) (\frac{1 - \cos^2 2x}{4}) \\ & = & 4(\cos^3 2x - \cos^2 2x - \cos 2x + 1) \\ \end{aligned}$

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$\because \cos 3x = 4 \cos^3 x - 3 \cos x \Rightarrow \cos^3 x = \frac{1}{4} (\cos 3x + 3 \cos x)$ $\boxed{\therefore \cos^3 2x = \frac{1}{4} (\cos 6x + 3 \cos 2x) \dashrightarrow 1} \\$ $\because \cos^2 x = \frac{1 + \cos 2x}{2} \\ \boxed{\therefore \cos^2 2x = \frac{1 + \cos 4x}{2} \dashrightarrow 2}$

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Substituting $1$ and $2 :$

$\begin{aligned} \therefore 32 \sin^4 x \cos^2 x & = & 4[\frac{1}{4} (\cos 6x + 3 \cos 2x) - (\frac{1 + \cos 4x}{2}) - \cos 2x + 1] \\ & = & \cos 6x + 3 \cos 2x - 2 - 2 \cos 4x - 4 \cos 2x + 4\end{aligned}$

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$\therefore 32 \sin^4 x \cos^2 x = \cos 6x - 2 \cos 4x - \cos 2x + 2 \\$

$\Rightarrow \boxed{a = 2}$

We will use the double angle formula $\sin 2A = 2 \sin A \cos A, \cos 2A = 1 - 2\sin^2 A$ and the product to sum formula $2 \cos P \cos Q = \cos \left ( \frac {P+Q}{2} \right ) \cos \left ( \frac {P-Q}{2} \right )$

$\begin{aligned} 32 \sin^4 x \cos^2 x & = & 4 \cdot 4\sin^2 x \cos^2 x \cdot 2 \sin^2 x \\ & = & 4 (\sin^2 2x )(1 - \cos 2x ) \\ & = & 2(1 - \cos 4x)(1 - \cos 2x) \\ & = & 2 \cos 4x \cos 2x - 2 \cos 2x - 2 \cos 4x + 2 \\ \cos 6x - 2\cos 4x - \cos 2x + a & = & 2 \cos 4x \cos 2x - 2 \cos 2x - 2 \cos 4x + 2 \\ a + \cos 6x & = & 2 \cos 4x \cos 2x - \cos 2x + 2 \\ & = & \cos 6x + \cos 2x - \cos 2x + 2 \\ a & = & \boxed{2} \\ \end{aligned}$