Recall the double angle identity for sine,
sin(2x) = 2 sin(x) cos(x)
as well as the definition of tangent,
tan(x) = sin(x)/cos(x)
Now rewrite the equation as
3 tan(x) = 2 sin(2x)
3 sin(x)/cos(x) = 4 sin(x) cos(x)
3 sin(x)/cos(x) - 4 sin(x) cos(x) = 0
sin(x)/cos(x) × (3 - 4 cos²(x)) = 0
We have two case:
• sin(x) = 0
⇒ x = arcsin(0) + 2nπ or x = π - arcsin(0) + 2nπ
⇒ x = 0 + 2nπ or x = π + 2nπ
(where n is any integer)
• 3 - 4 cos²(x) = 0
⇒ 4 cos²(x) = 3
⇒ cos²(x) = 3/4
⇒ cos(x) = ± √3/2
⇒ x = arccos(√3/2) + 2nπ or x = -arccos(√3/2) + 2nπ
… … … or x = arccos(-√3/2) + 2nπ or x = -arccos(-√3/2) + 2nπ
⇒ x = π/6 + 2nπ or x = -π/6 + 2nπ
… … … or x = 5π/6 + 2nπ or x = -5π/6 + 2nπ
We also observe that any x such that cos(x) = 0 cannot be a solution. But this doesn't happen for with any of the solutions found above.
In the interval [0, 2π), we get the solutions
x = 0, π/6, 5π/6, π, 7π/6, 11π/6