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

$({\sin }^2\theta +{\cos }^2\theta {)}^2-2{\sin }^2\theta {\cos }^2\theta -\sin \theta \cos \theta =0$
Let $\sin \theta \cdot \cos \theta =t,$
$\Rightarrow 1-2t^2-t=0$
$\Rightarrow 2t^2+t-1=0$
$\Rightarrow t=\frac{1}{2},-1$
$\Rightarrow \sin \theta \cos \theta =\frac{1}{2}\text{or}\sin \theta \cos \theta =-1$
$\Rightarrow \sin 2\theta =1\text{or}\sin 2\theta =-2\text{(Not Possible)}$
$\therefore \theta =\frac{\pi }{4},\frac{5\pi }{4},\frac{9\pi }{4},\frac{13\pi }{4}$
$\therefore S=7\pi$
$\Rightarrow \frac{8S}{\pi }=56$