Answers

2014-04-28T18:45:52+05:30
sin(  \frac{ \pi x}{2} )= x^{2} -2x+2
sin's range is [-1,1]
 1 \geq x^{2} -2x+2 \geq -1
so
 x^{2} -2x+3 \geq 0
for this discriminat is 4-12=-8
so for any real value of x is it always greater than zero
so
 x^{2} -2x+3>0 \\ x^2-2x+2>-1
then
1  \geq  x^2-2x+2 \\ 0 \geq  x^{2} -2x+1 \\ 0 \geq (x-1)^2
square of any number can't negative
so 
x-1=0 \\ x=1
so only for x=1 this relation holds
graph
 
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