The correct option is C y=(4n+1)π2, n∈I
→a×(→b×→b)+(→a⋅→b)→b =(4−2x−siny)→b+(x2−1)→c⇒(→a⋅→c)→b−(→a⋅→b)→c+(→a⋅→b)→b =(4−2x−siny)→b+(x2−1)→c⋯(i)
We know that,
(→c⋅→c)→a=→c⇒(→c⋅→c)→a⋅→c=→c⋅→c⇒→a⋅→c=1
1+→a⋅→b=4−2x−siny and x2−1=−(→a⋅→b)
1+1−x2=4−2x−siny⇒siny=x2−2x+2⇒siny=(x−1)2+1
We know that,
−1≤siny≤1 and
(x−1)2+1≥1
So the solution will be,
siny=1⇒y=(4n+1)π2, n∈I
and x=1