Question

sin ( pie x/2) = x^2 - 2x +2 , find x

Answer 1

$sin( \frac{ \pi x}{2} )= x^{2} -2x+2$
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
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