Given, √3cos2x=(√3−1)cosx+1,x∈[0,π∕2] Let cosx=t, then √3t2=(√3−1)t+1 ⇒√3t2−√3t+t−1=0 ⇒(√3t2−√3t)+(t−1)=0 ⇒√3t(t−1)+1(t−1)=0 ⇒(t−1)(√3t+1)=0 This gives t=1 and t=
−1
√3
Put, t=cosx, then cosx=1 and cosx=
−1
√3
cosx=−1∕√3 is rejected as x∈[0,π∕2] cosx=−1∕√3 is rejected as x∈[0,π ∴cosx=1 Since, x∈[0,
π
2
], then cosx=cos0 This gives x=0 is only solution. Therefore, number of solution when x∈[0,π∕2] is 1 .
