Question 891651
{{{graph(300,300,-10,10,-10,10,sin(pi*x/2),x^2-2x+2)}}}
.
.
.
{{{graph(300,300,-2,2,-2,2,sin(pi*x/2),x^2-2x+2)}}}
.
.
.
Looks like there is a solution at {{{x=1}}}.
Verify.
{{{sin(pi/2)=1}}}
{{{(1)^2-2(1)+2=1-2+2=1}}}
Yes, ({{{1}}},{{{1}}}) is an intersection point of the two curves.