If the equation sin ( π x 2 ) – sin( π x 2 + 2 π x) = 0 is solved for positive roots, then in the increasing sequence of positive root
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(b,c)
⇒ 
Case- I n is even ⇒ n = 2m
then x 2 = 2m + x 2 +2x ⇒ x = – m
This gives positive root as 1, 2, 3, .....corresponding to m = – 1, –2, – 3, ......
Case-II n is odd ⇒ n = 2m+1
⇒ x 2 = (2m+1) – (x 2 +2x)
⇒ x = 
This gives poistive roots 
, 
, 
.....
⇒ positive roots in increasing sequence are
, 
, 1, 
, 
, 
, 
,2, .....
⇒ 8 term of sequence is 2
(b,c)
⇒ 
Case- I n is even ⇒ n = 2m
then x 2 = 2m + x 2 +2x ⇒ x = – m
This gives positive root as 1, 2, 3, .....corresponding to m = – 1, –2, – 3, ......
Case-II n is odd ⇒ n = 2m+1
⇒ x 2 = (2m+1) – (x 2 +2x)
⇒ x = 
This gives poistive roots 
, 
, 
.....
⇒ positive roots in increasing sequence are
, 
, 1, 
, 
, 
, 
,2, .....
⇒ 8 term of sequence is 2
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