We have, ∣sinxcosxcosxcosxsinxcosxcosxcosxsinx∣=0,−4π≤x≤4π
Apply: R1→R1−R2 and R2→R2−R3
∣sinx−cosx0cosxcosx−sinxsinx−cosxcosx0cosx−sinxsinx∣=0
⇒(sinx−cosx)2∣10cosx−11cosx0−1sinx∣=0
⇒(sinx−cosx)2(sinx+2cosx)=0
⇒sinx=cosx or sinx=−2cosx
∴x=4π
Since, 4−π≤x≤4π, we get −1≤tanx≤1
⇒tanx=1⇒x=4π
Hence, number of roots is one.