Question

The principal solutions of tan 30 = -1 are

• A. $\left\{\frac{\pi}{4}, \frac{7\pi}{12}, \frac{11\pi}{12}, \frac{\pi}{16}, \frac{19\pi}{4}, \frac{23\pi}{24}\right\}$
• B. $\left\{\frac{\pi}{4}, \frac{7\pi}{12}, \frac{11\pi}{12}, \frac{5\pi}{4}, \frac{19\pi}{12}, \frac{23\pi}{12}\right\}$
• C. $\left\{\frac{\pi}{4}, \frac{\pi}{12}\right\}$
• D. $\left\{\frac{\pi}{4}, \frac{\pi}{12}, \frac{13\pi}{12}, \frac{7\pi}{4}, \frac{19\pi}{4}, \frac{23\pi}{12}\right\}$

Answer 1

Answer: (B)

$\left\{\frac{\pi}{4}, \frac{7\pi}{12}, \frac{11\pi}{12}, \frac{5\pi}{4}, \frac{19\pi}{12}, \frac{23\pi}{12}\right\}$

Answer 2

The given equation is tan(3θ) = -1.

The general solution for tan(x) = -1 is given by x = nπ - π/4, where n is an integer.

So, 3θ = nπ - π/4.

Dividing by 3, we get θ = nπ/3 - π/12.

We need to find the principal solutions, which lie in the interval [0, 2π).

Let's substitute integer values for n and find the values of θ in the required interval:

For n = 1, θ = π/3 - π/12 = 4π/12 - π/12 = 3π/12 = π/4. (0 ≤ π/4 < 2π)

For n = 2, θ = 2π/3 - π/12 = 8π/12 - π/12 = 7π/12. (0 ≤ 7π/12 < 2π)

For n = 3, θ = 3π/3 - π/12 = π - π/12 = 12π/12 - π/12 = 11π/12. (0 ≤ 11π/12 < 2π)

For n = 4, θ = 4π/3 - π/12 = 16π/12 - π/12 = 15π/12 = 5π/4. (0 ≤ 5π/4 < 2π)

For n = 5, θ = 5π/3 - π/12 = 20π/12 - π/12 = 19π/12. (0 ≤ 19π/12 < 2π)

For n = 6, θ = 6π/3 - π/12 = 2π - π/12 = 24π/12 - π/12 = 23π/12. (0 ≤ 23π/12 < 2π)

For n = 7, θ = 7π/3 - π/12 = 28π/12 - π/12 = 27π/12 = 9π/4. (9π/4 = 2π + π/4, which is outside [0, 2π))

The principal solutions are {π/4, 7π/12, 11π/12, 5π/4, 19π/12, 23π/12}.