Answer:
Option C
Explanation:
Given, √3secx+cosecx+2(tanx−cotx)=0,
(−π<x<π)−(0,±π/2)
⇒ √3sinx+cosx+2(sin2x−cos2x)=0
⇒ √3sinx+cosx−2cos2x=0
Multiplying and dividing by √a2+b2. i.e
√3+1=2 , we get
2(√32sinx+12cosx)−2cos2x=0
= (cosx.cosπ3+sinx.sinπ3)−cos2x=0
cos(x−π3)=cos2x
∴ ∴2x=2nπ±(x−π3)
[since., cosθ=cosα⇒θ=2nπ±α]
⇒ 2x=2nπ+x−π3
or 2x=2nπ−x+π3
⇒x=2nπ−π3or3x=2nπ+π3
⇒x=2nπ−π3orx=2nπ3+π9
∴ x=−π3 or x=π9.−5π9.7π9
Now, sum of all distinct solutions
=−π3+π9−5π9+7π9=0