wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

If ∣ ∣ ∣2cos2xcos2x8sin2xcos2xsin2x2sin2x8sin2xcos2xsin2xcos2x1+4sin4x∣ ∣ ∣=0where x(0,π),
then the number of solutions is

Open in App
Solution

C1C1+C2+C3∣ ∣ ∣2+4sin4xcos2x4sin4x2+4sin4x2sin2x4sin4x2+4sin4xcos2x1+4sin4x∣ ∣ ∣=0(2+4sin4x)∣ ∣ ∣1cos2x4sin4x11+cos2x4sin4x1cos2x1+4sin4x∣ ∣ ∣=0R1R1R3, R2R2R3(2+4sin4x)∣ ∣0010111cos2x1+4sin4x∣ ∣=0(2+4sin4x)×1=0sin4x=12=sinπ6sin4x=sin(π+π6) or sin(2ππ6)4x=2πn+7π6 or 4x=2πn+11π6x=πn2+7π24 or x=πn2+11π24x=7π24,11π24,19π24,23π24Number of solutions is 4

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon