Question

The total number of solution of $si{n}^{4}x+co{s}^{4}x=sinxcosx$ in $[0,2\pi ]$ is equal to ______

Answer 1

Answer: 2

$si{n}^{4}x+co{s}^{4}x=sinxcosx$
or $(si{n}^{2}x+co{s}^{2}x{)}^2-2si{n}^{2}xco{s}^{2}x=sinxcosx$
or $1-\frac{si{n}^{2}2x}{2}=\frac{sin2x}{2}$
or $si{n}^{2}2x+sin2x-2=0$
or $(sin2x+2)(sin2x-1)=0$
or $sin2x=1$
or $2x=(4n+1)\frac{\pi }{2},n\in Z$
or$x=(4n+1)\frac{\pi }{4},n\in Z$
$=\frac{\pi }{4},\frac{5\pi }{4}(\because x\in [0,2\pi ])$
Thus, there are two solutions.