The number of solutions of the equation|cot x| = cot x + 1sinx ,0 < x < 2π, is
1
If cot x > 0
then 1sinx = 0 (impossible) Now if cot x < 0, then
-cot x = cot x + 1sinx
⇒2cosx+1sinx = 0 ⇒ cos x = - 12
⇒ cos x = cos( 2π3)
x = 2nπ ± 2π3; n ∈ I
and 0 ≤ x ≤ 2π then x = 2π3, 4π3
But cotx < 0 ∴ x = 2π3