Correct Answer - C
Give `(sqrt(3)) sec x + coses x + 2 (tan x - cot c) = 0`
`(-pi lt x lt pi) - {0,pm pi//2}`
`rArr sqrt(3) x cos x + 2 (sin^(2) x - cos^(2)x) = 0`
`rArr sqrt(3) sin x + cos x - 2 cos 2 x =0`
Multiplying and dividing by `sqrt(a^(2) + b^(2)) i.e., sqrt(3+1) = 2`.
We get
`2((sqrt(3))/(2) sin x + (1)/(2)cosx)- 2 cps 2x = 0`
`rArr (cos x cos .(pi)(3) + sin x.sin.(pi)/(2))-cos 2 x cos = 0 cos (x-(pi)/(3)) = cos2x`
`therefore " " 2x = 2npi+x -(pi)/(3)" " [underset(rArr theta = 2npipmalpha)"Since",cos theta =cos alpha] `
`rArr 2x = 2npi + x- (pi)/(3)`
`or " " 3x = 2xpi-x + (pi)/(3)`
`rArr" "x = 2npi - (pi)/(3)`
`or " " 3x = 2npi + (pi)/(3)`
`rArr x = 2npi - (pi)/(3)`
`or " " x= (2npi)/(3) + (pi)/(9)`
`therefore" " x= (-pi)/(3)`
` or" "x=(pi)/(9),(-5pi)/(9) ,(7pi)/(9)`
Now, sum of all distinct solutions `= (-pi)/(3) + (pi)/(9)-(5pi)/(9) + (7pi)/(9) = 0`
