Question 696709
Let 
{{{ y = sqrt(-x^2+4x-3) + sqrt(sin(pi/2*sin(pi/2)(x-1) )) }}}


Then factorising the first term and tidying up the second gives:
{{{ y = sqrt((-x+1)(x-3)) + sqrt(sin(pi/2*(x-1) )) }}}



As we are assuming a real-valued function then both terms under the square root signs must be simultaneously positive (or zero).



Firstly consider {{{ (-x+1)(x-3) >= 0 }}}
{{{ graph( 200, 200, -1, 4, -10, 10, -x^2+4x-3 ) }}}
The solution to this is: {{{ 1 <= x <=3 }}}



Secondly, consider {{{ sin(pi/2*(x-1)) >= 0}}}
{{{ graph( 200, 200, -2pi, 2pi, -3, 3, sin(pi/2*(x-1)) ) }}}
{{{ sin(pi/2*(x-1)) >= 0 }}}
{{{ 0<= pi/2*(x-1) <= pi }}}
{{{ 0<= x-1 <= 2 }}}
{{{ 1<= x <= 3 }}}



So in both cases, the square root terms simultaneously exist when {{{ 1<= x <= 3 }}}, and this is the range.