if a be the smallest positive root of the equation ?sin(1-x) = ?cosx, then the approximate integral value of (a-1) must be?
?sin(1-x) = ?cosx
Squaring both side with condition that sin(1-x)>0,cosx>0
Considering only negative sign as +sign will give no solution
For smallest positive root
x=3?/4+1/2 does not satisfy the condition that sin(1-x)>0,cosx>0
x=7?/4+1/2satisfy the condition that sin(1-x)>0,cosx>0
so approximate integral value of root=6