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

Solve rsinθ=3,r=4(1+sinθ),0θ2π.

Open in App
Solution

Eliminating r, we get 4sin2θ+4sinθ3=0
or (2sinθ+3)(2sinθ1)=0
sinθ=12=sinπ6
θ=π/6,ππ/6 for 0θ2π.
The other value is rejected as sinθ cannot be greater than 1.

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