b→ and c→ are non-collinear If a→×(b→×c→)+(a→⋅b→)b→=(4−2x−siny)b→+x2−1c→ and (c→⋅c→)a→=c→ then
x=1
x=-1
y=(4n+1)π2,n∈I
y=(2n+1)π2,n∈I
a→×(b→×c→)+(a→⋅b→)b→=(4−2x−siny)b→+x2−1c→ or (a→⋅c→)b→−(a→⋅b→)c→+(a→⋅b→)b→=(4−2x−siny)b→+x2−1c→
now , (c→⋅c→)a→=c→. Therefore
(c→⋅c→)(a→⋅c→)=(c→⋅c→) or a→⋅c→=1⇒1+a→⋅b→=4−2x−siny,x2−1=−(a→⋅b→)or 1=4−2x−siny+x2−1or siny=x2−2x+2=(x−1)2+1but siny≤1⇒x=1,siny=1⇒y=(4n+1)π2, n∈I