Total number of solutions of ${\sin }^4\mathrm{x}+{\cos }^4\mathrm{x}=\sin \mathrm{x}\cdot \cos \mathrm{x}\text{ in }[0,2\mathrm{\pi }]$ is equal to

Given,${\sin }^4\mathrm{x}+{\cos }^4\mathrm{x}=\sin \mathrm{x}\cdot \cos \mathrm{x}$ 

$\begin{array}{ll}\Rightarrow & {\left({\sin }^2\mathrm{x}+{\cos }^2\mathrm{x}\right)}^2-2{\sin }^2\mathrm{x}\cdot {\cos }^2\mathrm{x}=\sin \mathrm{x}\cdot \cos \mathrm{x}\\ \Rightarrow & 1-\frac{{\sin }^{2}2\mathrm{x}}{2}=\frac{\sin 2\mathrm{x}}{2}\\ \Rightarrow & {\sin }^{2}2\mathrm{x}+\sin 2\mathrm{x}-2=0\\ \Rightarrow & (\sin 2\mathrm{x}+2)(\sin 2\mathrm{x}-1)=0\\ \Rightarrow & \sin 2\mathrm{x}=1\\ \therefore & 2\mathrm{x}=(4\mathrm{n}+1)\frac{\mathrm{\pi }}{2}\\ \Rightarrow & \mathrm{x}=(4\mathrm{n}+1)\frac{\mathrm{\pi }}{4}\Rightarrow \mathrm{x}=\frac{\mathrm{\pi }}{4},\frac{5\mathrm{\pi }}{4}\end{array}$

Hence, two solutions exist.