X. sin(πcosx)=0 needs cosx∈{−1,0,1}, so X={2kπ}. That is an arithmetic progression and contains 2π,23π,4π,7π.
Y.
f′(x)=−πsinxcos(πcosx)=0⟺sinx=0 or cosx=±21.
So Y={3kπ}. That is an arithmetic progression and contains 3π,32π,π.
The options.
- (A) fails because X is an AP, so not (R).
- (B) (II), (Q), (T) is correct.
- (C) fails: X does not contain 6π.
- (D) fails: Y is an AP.
(B)