Question

Solve the following equation in the interval [0, 2 pi].

Note: Give the answer as a multiple of pi. Do not use decimal numbers. The answer should be a fraction or an integer. Note that pi is already included in the answer so you just have to enter the appropriate multiple. E.g. if the answer is pi/2 you should enter 1/2. If there is more than one answer enter them separated by commas
|tan(t)| = 1/sqrt(3)

Answer 1

The solution of the following equation |tan(t)| = 1/√3 in the interval [0, 2 pi]

t = (π/3, 2π/3, 4π/3, 5π/3)

To solve the following equation |tan(t)| = 1/√3 in the interval [0, 2 pi], we proceed as follows

The trigonometric equation in modulus form |tan(t)| = 1/sqrt(3) implies that

tan(t) = 1/√3 and -tan(t) = 1/√3

So, solving these, we have that

tan(t) = 1/√3 and -tan(t) = 1/√3

tan(t) = 1/√3 and tan(t) = -1/√3

We know that tan(t) is positive in the first and third quadrant and negative in the second and fourth quadrant. So, we have that taking inverse

tan(t) = 1/√3 or tan(3π/2 - t) = 1/√3 and tan(π - t) = 1/√3 or tan(2π - t) = 1/√3

Taking inverse tan, we have that

t = tan⁻¹(1/√3) or tan(3π/2 - t) = tan⁻¹(1/√3) and π - t = tan⁻¹(1/√3) or 2π - t = tan⁻¹(1/√3)

t = π/3 or π + t = π/3 and π - t = π/3 or 2π - t = π/3

t = π/3 or -t = π - π/3 and t = π - π/3 or t = 2π - π/3

t = π/3 or -t = (3π - π)/3 and t = (3π - π)/3 or t = (6π - π)/3

t = π/3 or -t = 2π/3 and t = 2π/3 or t = 5π/3

t = π/3 or t = -2π/3 and t = 2π/3 or t = 5π/3

t = π/3 or t = -2π/3 + 2π and t = 2π/3 or t = 5π/3

t = π/3 or t = (-2π + 6π)/3 and t = 2π/3 or t = 5π/3

t = π/3 or t = 4π/3 and t = 2π/3 or t = 5π/3

So, t = (π/3, 2π/3, 4π/3, 5π/3)