The correct option is C 1
Given, ∣∣
∣∣sin xcos xcos xcos xsin xcos xcos xcos xsin x∣∣
∣∣=0
Applying C1→C1+C2+C3
=∣∣
∣∣sin x+2 cos xcos xcos xsin x+2 cos xsin xcos xsin x+2 cos xcos xsin x∣∣
∣∣
=(2 cos x+sin x)∣∣
∣∣1cos xcos x1sin xcos x1cos xsin x∣∣
∣∣=0
Applying R2→R2−R1, R3→R3−R1
⇒(2 cos x+sin x)∣∣
∣∣1cos xcos x0sin x−cos x000sin x−cos x∣∣
∣∣=0⇒ (2 cos x+sin x)(sin x−cos x)2=0
⇒ 2 cos x + sin x = 0 or sin x - cos x = 0
⇒ 2 cos x =-sin x or sin x = cos x
⇒ cot x=−12 gives no solution in −π4≤x≤π4
and sin x = cos x ⇒ tan x = 1 ⇒ x=π4