If sin 6 theta = 32 cos^5 theta sin theta - 3 | vedclass

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(D) We know that $\sin 6 	heta = \sin 3(2 	heta)$.Using the identity $\sin 3A = 3 \sin A - 4 \sin^3 A$,we get:$\sin 6 	heta = 3 \sin 2 	heta - 4 \

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$\sin 2 	heta$

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(D) We know that $\sin 6 	heta = \sin 3(2 	heta)$.Using the identity $\sin 3A = 3 \sin A - 4 \sin^3 A$,we get:$\sin 6 	heta = 3 \sin 2 	heta - 4 \sin^3 2 	heta$.Substitute $\sin 2 	heta = 2 \sin 	heta \cos 	heta$:$\sin 6 	heta = 3(2 \sin 	heta \cos 	heta) - 4(2 \sin 	heta \cos 	heta)^3$$= 6 \sin 	heta \cos 	heta - 4(8 \sin^3 	heta \cos^3 	heta)$$= 6 \sin 	heta \cos 	heta - 32 \sin^3 	heta \cos^3 	heta$$= 6 \sin 	heta \cos 	heta - 32 \sin 	heta \cos^3 	heta (1 - \cos^2 	heta)$$= 6 \sin 	heta \cos 	heta - 32 \sin 	heta \cos^3 	heta + 32 \sin 	heta \cos^5 	heta$$= 32 \cos^5 	heta \sin 	heta - 32 \cos^3 	heta \sin 	heta + 3(2 \sin 	heta \cos 	heta)$$= 32 \cos^5 	heta \sin 	heta - 32 \cos^3 	heta \sin 	heta + 3 \sin 2 	heta$.Comparing this with the given equation $\sin 6 	heta = 32 \cos^5 	heta \sin 	heta - 32 \cos^3 	heta \sin 	heta + 3x$,we find $3x = 3 \sin 2 	heta$,which implies $x = \sin 2 	heta$.

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

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