CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

If xϵ[0,π] and 3sin2x+2cos2x+31sin2x+2sin2x=28, then find number of values of x satisfy given equation.

Open in App
Solution

3sin2x+2sin2x+31sin2x+2sin2x=283sin2x+2cos2x+33(sin2x+2cos2x)=28
Let y=3sin2x+2cos2x
y=27y=28
y228y+27=0y=27 or 1
If y=27 then 3sin2x+2cos2x=33
sin2x+(2cos2x1)=0
sin2x+cos2x=2
Dividing both sides by 2
sin(2x+π4)=2
which is not possible
Now if y=1
3sin2x+2cos2x=1=30sin2x+2cos2x=02cosx(sinx+cosx)=0
cosx=0 or tanx=1

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