If sin 6theta = 32cos^5theta sintheta - 32cos | vedclass

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(D) We know that $\sin 6	heta = 2\sin 3	heta \cos 3	heta$. Using the identities $\sin 3	heta = 3\sin	heta - 4\sin^3	heta$ and $\cos 3	heta = 4\

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$\sin 2	heta$

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(D) We know that $\sin 6	heta = 2\sin 3	heta \cos 3	heta$. Using the identities $\sin 3	heta = 3\sin	heta - 4\sin^3	heta$ and $\cos 3	heta = 4\cos^3	heta - 3\cos	heta$: $\sin 6	heta = 2(3\sin	heta - 4\sin^3	heta)(4\cos^3	heta - 3\cos	heta)$ $= 2(12\sin	heta \cos^3	heta - 9\sin	heta \cos	heta - 16\sin^3	heta \cos^3	heta + 12\sin^3	heta \cos	heta)$ $= 24\sin	heta \cos^3	heta - 18\sin	heta \cos	heta - 32\sin^3	heta \cos^3	heta + 24\sin^3	heta \cos	heta$ By rearranging and using $\sin^2	heta = 1 - \cos^2	heta$,we simplify the expression to match the form $32\cos^5	heta \sin	heta - 32\cos^3	heta \sin	heta + 3x$. After algebraic simplification,we find that $3x = 3\sin 2	heta$. Therefore,$x = \sin 2	heta$.

If $\sin 6\theta = 32\cos^5\theta \sin\theta - 32\cos^3\theta \sin\theta + 3x$,then $x = $

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