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

Find all solutions of 4cos2xsinx2sin2x=3sinx

A
x=2nπ orx=mπ±(1)msin1(1±54)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
x=nπ orx=2mπ±(1)msin1(1±54)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
x=nπ orx=mπ±(1)msin1(1±54)
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
none of these
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is D x=nπ orx=mπ±(1)msin1(1±54)
4cos2xsinx2sin2x=3sinx4(1sin2x)sinx2sin2x=3sinx4sinx4sin3x2sin2x=3sinx4sin3x2sin2x+sinx=04sin3x+2sin2xsinx=0sinx(4sin2x+2sinx1=0)
sinx=0 or 4sin2x+2sinx1=0
sinx=0 or sinx=2±4+1624
sinx=0 or sinx=1±54
x=nπ or x=mπ+(1)msin1(1±54)

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