The number of Solutions of the equation $Sin2x=[1+\sin 2x]+[1-\cos 2x]$
  in $[0,\frac{\pi }{2}]$  is __________  (where [.] is G.I.F)
 

 $[1-\cos 2x]=\left[2{\sec }^{2}x\right]$
 
 $[1+\sin 2x]=[1+\cos (\frac{\pi }{2}-2x)]=\left[2{\cos }^2(x-\frac{\pi }{4})\right]$
 $\therefore \sin 2x=\left[2{\cos }^2(x-\frac{\pi }{4})\right]+\left[2{\sin }^{2}x\right]$
       $\left[2{\sin }^{2}x\right]=0ifx\in [0,\frac{\pi }{4})$
 $=1ifx\in (\frac{\pi }{4},\frac{\pi }{2})$
 $\left[2{\cos }^2(x-\frac{\pi }{4})\right]=1ifx\in [0,\frac{\pi }{4})$      
 $=1ifx\in [\frac{\pi }{4},\frac{\pi }{2}]$
  $\sin 2x=1ifx\in [0,\frac{\pi }{4})$
 $=2ifx\in [\frac{\pi }{4},\frac{\pi }{2}]$
$\sin 2x=1,x=\frac{\pi }{4}\notin [0,\frac{\pi }{4})$

There is no x