Question

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_____

Answer 1

Answer: 56

Given equation
${\sin }^4\theta +{\cos }^4\theta -\sin \theta \cos \theta =0$
$\Rightarrow 1-{\sin }^2\theta {\cos }^2\theta -\sin \theta \cos \theta =0$
$\Rightarrow 2-(\sin 2\theta {)}^2-\sin 2\theta =0$
$\Rightarrow (\sin 2\theta {)}^2+(\sin 2\theta )-2=0$
$\Rightarrow (\sin 2\theta +2)(\sin 2\theta -1)=0$
$\Rightarrow \sin 2\theta =1$ or $\sin 2\theta =-2$ (not possible)
$\Rightarrow 2\theta =\frac{\pi }{2},\frac{5\pi }{2},\frac{9\pi }{2},\frac{13\pi }{2}$
$\Rightarrow \theta =\frac{\pi }{4},\frac{5\pi }{4},\frac{9\pi }{4},\frac{13\pi }{4}$
$\Rightarrow S=\frac{\pi }{4}+\frac{5\pi }{4}+\frac{9\pi }{4}+\frac{13\pi }{4}=7\pi$
$\Rightarrow \frac{8S}{\pi }=\frac{8\times 7\pi }{\pi }=56.00$