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

Consider the system of equations
xcos3y+3xcosysin2y=14
xsin3y+3xcos2ysiny=13

The number of values of y[0,6π] is

A
5
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
3
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
4
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
6
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct option is D 6
xcos3y+3xcosysin2y=14 (1)
xsin3y+3xcos2ysiny=13 (2)
Adding Eqs. (1) and (2)
x(cos3y+3cos2ysiny+3cosysin2y+sin3y)=27
x(cosy+siny)3=27
x1/3(cosy+siny)=3 (3)

Subtracting Eq (2) from Eq (1), we get
x(cos3y+3cosysin2y3cos2ysinysin3y)=1
x(cosysiny)3=1
x1/3(cosysiny)=1 (4)

Dividing Eq (3) by Eq (4), we get
cosy+siny=3cosy3siny
tany=12

tan function is positive in 1st quadrant and 3rd quadrant.
Therefore, there are exactly six values of y[0,6π]

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