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

If x+y=2π3 and sinxsiny=2, then which of the following option(s) is/are correct?

A
there are 4 values of x in [0,4π] for yR{nπ}
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
there are 6 values of x in [0,4π] for yR{nπ}
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
there are 6 values of y in [0,4π] for xR
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
there are 4 values of y in [0,4π] for xR
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct options are
A there are 4 values of x in [0,4π] for yR{nπ}
D there are 4 values of y in [0,4π] for xR
x+y=2π3y=2π3x
sinx=2sin(2π3x)
sinx=2[(32)cosx+12sinx]sinx=3cosx+sinxcosx=0x=(2n+1)π2, nZ
y=2π3nππ2
y=π6nπ, nZ

Hence, for x[0,4π],
x=π2,3π2,5π2,7π2

and for y[0,4π],
y=π6,7π6,13π6,19π6

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon