Let S be the sum of all solutions (in radians | vedclass

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(C) Given equation is $\sin^{4} 	heta + \cos^{4} 	heta - \sin 	heta \cos 	heta = 0$.We know that $\sin^{4} 	heta + \cos^{4} 	heta = (\sin^{2}

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$56$

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(C) Given equation is $\sin^{4} 	heta + \cos^{4} 	heta - \sin 	heta \cos 	heta = 0$.We know that $\sin^{4} 	heta + \cos^{4} 	heta = (\sin^{2} 	heta + \cos^{2} 	heta)^{2} - 2\sin^{2} 	heta \cos^{2} 	heta = 1 - 2\sin^{2} 	heta \cos^{2} 	heta$.Substituting this into the equation: $1 - 2\sin^{2} 	heta \cos^{2} 	heta - \sin 	heta \cos 	heta = 0$.Using $2\sin 	heta \cos 	heta = \sin 2	heta$,we have $\sin^{2} 	heta \cos^{2} 	heta = \frac{\sin^{2} 2	heta}{4}$.So,$1 - 2(\frac{\sin^{2} 2	heta}{4}) - \frac{\sin 2	heta}{2} = 0$.Multiplying by $2$: $2 - \sin^{2} 2	heta - \sin 2	heta = 0$,which simplifies to $\sin^{2} 2	heta + \sin 2	heta - 2 = 0$.Factoring gives $(\sin 2	heta + 2)(\sin 2	heta - 1) = 0$.Since $\sin 2	heta$ cannot be $-2$,we have $\sin 2	heta = 1$.For $	heta \in [0, 4\pi]$,$2	heta \in [0, 8\pi]$.Thus,$2	heta = \frac{\pi}{2}, \frac{5\pi}{2}, \frac{9\pi}{2}, \frac{13\pi}{2}$.$	heta = \frac{\pi}{4}, \frac{5\pi}{4}, \frac{9\pi}{4}, \frac{13\pi}{4}$.The sum $S = \frac{\pi + 5\pi + 9\pi + 13\pi}{4} = \frac{28\pi}{4} = 7\pi$.Therefore,$\frac{8S}{\pi} = \frac{8(7\pi)}{\pi} = 56$.

Let $S$ be the sum of all solutions (in radians) of the equation $\sin^{4} \theta + \cos^{4} \theta - \sin \theta \cos \theta = 0$ in $[0, 4\pi]$. Then $\frac{8S}{\pi}$ is equal to ...... .

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