Question

Total number of solutions for the equation ${\sin }^{4}x+{\cos }^{4}x=\sin x\cos x,x\in [0,2\pi ]$ is

• A. $4$
• B. $2$
• C. $6$
• D. $3$

Answer 1

The correct option is B $2$

${\sin }^{4}x+{\cos }^{4}x=\sin x\cos x\Rightarrow ({\sin }^{2}x+{\cos }^{2}x{)}^2-2{\sin }^{2}x{\cos }^{2}x=\sin x\cos x\Rightarrow 1-\frac{{\sin }^{2}2x}{2}=\frac{\sin 2x}{2}\Rightarrow {\sin }^{2}2x+\sin 2x-2=0\Rightarrow (\sin 2x+2\left)\right(\sin 2x-1)=0\Rightarrow \sin 2x=1(\because -1\le \sin 2x\le 1)\Rightarrow 2x=(4n+1)\frac{\pi }{2},n\in Z\Rightarrow x=(4n+1)\frac{\pi }{4},n\in Z$
So, the solutions in $[0,2\pi ]$ are $x=\frac{\pi }{4},\frac{5\pi }{4}$