If −π/4≤x≤π/4,then the number of distinct real roots of sinxcosxcosxcosxsinxcosxcosxcosxsinx=0 is
see full answer
Your Exam Success, Personally Taken Care Of
1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya
a
0
b
2
c
1
d
3
answer is C.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
Using C1→C1+C2+C3,Δ=sinx+2cosxcosxcosxsinx+2cosxsinxcosxsinx+2cosxcosxsinx=(sinx+2cosx)1cosxcosx1sinxcosx1cosxsinxApplying R2→R2−R1 and R3→R3−R1, we getΔ=(sinx+2cosx)1cosxcosx0sinx−cosx000sinx−cosx =(sinx+2cosx)(sinx−cosx)2Thus, Δ=0⇒tanx=−2 or tanx=1As −π/4≤x≤π/4, we get −1≤tanx≤1∴ tanx=1⇒x=π/4