Given equation ∣cosxsinxsinxsinxcosxsinxsinxsinxcosx∣=0
Performing row and column operations
R1→R1−R2
R2→R2−R3
⇒∣cosx−sinx0sinxsinx−cosxcosx−sinxsinx0sinx−cosxcosx∣=0
C2→C2+C3
⇒∣cosx−sinx0sinxsinx−cosx0sinx+cosx0sinx−cosxcosx∣=0
Expanding using second row, we get
(sinx+cosx)(cosx−sinx)−sinx(sinx−cosx)=0
⇒(cosx−sinx)(sinx+cosx+sinx)=0
⇒(cosx−sinx)(2sinx+cosx)=0
⇒(cosx−sinx)=0
sinx=cosx⇒4π
Or
2sinx+cosx=0
tanx=−21
x=tan−1(−21)
Hence, 2 solutions are there.