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3 tan(x) = 2 sin(2x) from [0,2π)

User Jychan
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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

User Qiong
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