Question

The number of solutions of the equation ${\sin }^{4}x+{\cos }^{4}x=$ $\sin x\cos x$ in $[\pi ,5\pi ]$ is

Answer 1

${\sin }^{4}x+{\cos }^{4}x=\sin x\cos x$
$\Rightarrow {({\sin }^{2}x+{\cos }^{2}x)}^2-2{\sin }^{2}x{\cos }^{2}x=\sin x\cos x$
$\Rightarrow 2{\sin }^{2}x{\cos }^{2}x+\sin x\cos x-1=0$ $\Rightarrow (2\sin x\cos x-1)(\sin x\cos x+1)=0$
$\therefore \sin 2x=1[\because \sin x\cos x+1\ne 0]$

As $2x\in [2\pi ,10\pi ],$
$2x=\frac{5\pi }{2},\frac{9\pi }{2},\frac{13\pi }{2},\frac{17\pi }{2}$
$\therefore$ The number of solutions is $4$