Rewrite the right side as √3/3 = 1/√3, and recall that tan(x) = 1/√3 when x = π/6. Then since tan is π-periodic, taking the inverse tan of both sides gives
[tex]\tan\left(x + \dfrac\pi3\right) = \dfrac1{\sqrt3} \implies \tan^{-1}\left(\tan\left(x + \dfrac\pi3\right)\right) = \tan^{-1}\left(\dfrac1{\sqrt3}\right) + n\pi[/tex]
[tex]\implies x + \dfrac\pi3 = \dfrac\pi6 + n\pi[/tex]
where n is any integer. Solving for x, we get
[tex]x = -\dfrac\pi6 + n\pi[/tex]
and the solutions in the interval [0, 2π] are x = 5π/6 and x = 11π/6 (for n = 1 and n = 2).
