For non-collinear vectors b and c,|c|≠0, it is given that, (c⋅c)a=c⇒a⋅c=1 ...(i) and a×(b×c)+(a⋅b)b =(4−2β−sinα)b+(β2−1)c
⇒(a⋅c)b−(a⋅b)c+(a⋅b)b=(4−2β−sinα)b+(β2−1)c
So, a⋅c+a⋅b=4−2β−sinα....(ii) and −a⋅b=β2−1.....(ii) From Eqs. (i), (ii) and (iii), we get 1=4−2β−sinα+β2−1 ⇒(β−1)2−sinα=−1 So, if β=1, then sinα=1⇒α=2nπ+