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 $\frac{8 S}{\pi}$ is equal to_____

Answer 1

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)
$\Rightarrow 2 \theta=\frac{\pi}{2}, \frac{5 \pi}{2}, \frac{9 \pi}{2}, \frac{13 \pi}{2}$
$\Rightarrow \theta=\frac{\pi}{4}, \frac{5 \pi}{4}, \frac{9 \pi}{4}, \frac{13 \pi}{4}$
$\Rightarrow S =\frac{\pi}{4}+\frac{5 \pi}{4}+\frac{9 \pi}{4}+\frac{13 \pi}{4}=7 \pi$
$\Rightarrow \frac{8 S }{\pi}=\frac{8 \times 7 \pi}{\pi}=56.00$