Let S be the sum of all solutions (in radians) of the equation sin 4 θ + cos 4 θ - sin θ cos…

Given that: ${\sin }^4\theta +{\cos }^4\theta -\sin \theta \cos \theta =0\hspace{0.25em}\hspace{0.25em};\hspace{0.25em}\hspace{0.25em}\theta \in [0,4\pi ]$

$\Rightarrow 1-2{\sin }^2\theta {\cos }^2\theta -\sin \theta \cos \theta =0$

$\Rightarrow 2-{\sin }^{2}2\theta -\sin 2\theta =0$

$\Rightarrow {\sin }^{2}2\theta +\sin 2\theta -2=0$

$\Rightarrow \sin 2\theta =1 ;2\theta \in [0,8\pi ]$

$\theta =\frac{\pi }{4},\frac{5\pi }{4},\frac{9\pi }{4},\frac{13\pi }{4}$

$\Rightarrow$Sum of the solutions $=S=\frac{28\pi }{4}$
Then value of $\frac{8 \mathrm{S}}{\pi }=8\times \left(\frac{28\pi }{4\pi }\right)=56$