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