Solve tan x = √3 for x in [0, 2π).

Solve tan x = √3 for x in [0, 2π).

• A. x = π/4 and 3π/4
• B. x = π/3 and 4π/3
• C. x = π/3 and π/2
• D. x = 2π/3 and 5π/3

Answer: (B)

Tangents repeat every π, so solving tan x = √3 means finding angles where the acute reference angle has tangent √3. The standard angle with tan equals √3 is π/3, so all solutions are x = π/3 + kπ for integers k. Within [0, 2π), take k = 0 to get x = π/3, and k = 1 to get x = π/3 + π = 4π/3. No other k keeps you in the interval. Both of these angles lie in quadrants where tangent is positive, consistent with tan x = √3. So the solutions are x = π/3 and x = 4π/3.

Tangents repeat every π, so solving tan x = √3 means finding angles where the acute reference angle has tangent √3. The standard angle with tan equals √3 is π/3, so all solutions are x = π/3 + kπ for integers k.

Within [0, 2π), take k = 0 to get x = π/3, and k = 1 to get x = π/3 + π = 4π/3. No other k keeps you in the interval. Both of these angles lie in quadrants where tangent is positive, consistent with tan x = √3.

So the solutions are x = π/3 and x = 4π/3.