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

Sin (1-x)=cosx

Cosx=cos(?/2-1+x)

x=2n?±(?/2-1+x)n€I

Considering only negative sign as +sign will give no solution

x=2n?-?/2+1-x

2x=2n?-?/2+1

For smallest  positive root

n=1

2x=2?-?/2+1

x=3?/4+1/2 does not satisfy the condition that sin(1-x)>0,cosx>0

n=2

x=7?/4+1/2satisfy the condition that sin(1-x)>0,cosx>0

so approximate integral value of root=6

so a=6

a-1=5