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Basic Tangent Equations

A free Trigonometry lesson from the “Inverse Trig and Equations” unit, with a worked example and practice problems including step-by-step solutions.

A basic tangent equation tan(x) = k is solved by finding the principal angle from a known special value, then using tangent's period of pi to find every solution. Because tangent repeats every pi (not 2pi), each equation has exactly two solutions in [0, 2pi), found by adding pi to the first; the general solution is x = (principal angle) + n*pi.

What you'll learn

Why it matters: Periodic quantities that swing without bound between asymptotes, such as the slope of a rotating beam or the tangent of an elevation angle, are modeled by tangent equations whose repeated solutions come from the pi-length period.

Worked example

Problem. Solve tan(x) = sqrt(3) on the interval [0, 2pi).

  1. tan(pi/3) = sqrt(3), so the principal solution is x = pi/3.
  2. Tangent has period pi, so add pi: pi/3 + pi = 4pi/3.
  3. Both pi/3 and 4pi/3 lie in [0, 2pi).

Answer: x = pi/3, 4pi/3

Practice problems

1. Solve tan(x) = 1 on the interval [0, 2pi).

Show solution
  1. tan(pi/4) = 1, so the principal solution is x = pi/4.
  2. Add the period pi: pi/4 + pi = 5pi/4.
  3. Both values lie in [0, 2pi).

Answer: x = pi/4, 5pi/4

2. Solve tan(x) = 0 on the interval [0, 2pi).

Show solution
  1. Warm-up: First identify exactly what the question is asking: Solve tan(x) = 0 on the interval [0, 2pi).
  2. Use inverse operations to isolate the unknown, and keep both sides balanced at every step.
  3. Tangent is 0 when sine is 0, at x = 0.
  4. Add the period pi: 0 + pi = pi.
  5. Both 0 and pi are in [0, 2pi).
  6. Check the result by substituting or estimating: the response should match x = 0, pi and make sense in the original problem.

Answer: x = 0, pi

3. Solve tan(x) = sqrt(3) on the interval [0, 2pi).

Show solution
  1. tan(pi/3) = sqrt(3), so x = pi/3 is the principal solution.
  2. Add the period pi: pi/3 + pi = 4pi/3.
  3. Both values are in [0, 2pi).

Answer: x = pi/3, 4pi/3

4. Solve tan(x) = -1 on the interval [0, 2pi).

Show solution
  1. Tangent is negative in Quadrants II and IV; the reference angle is pi/4.
  2. In Quadrant II: pi - pi/4 = 3pi/4.
  3. Add pi for the Quadrant IV solution: 3pi/4 + pi = 7pi/4.

Answer: x = 3pi/4, 7pi/4

5. Solve tan(x) = sqrt(3)/3 on the interval [0, 2pi).

Show solution
  1. sqrt(3)/3 = 1/sqrt(3), and tan(pi/6) = sqrt(3)/3.
  2. So the principal solution is x = pi/6.
  3. Add the period pi: pi/6 + pi = 7pi/6.

Answer: x = pi/6, 7pi/6

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