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)

(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 general solution for $\tan x = -1$ is $x = n\pi + \frac{3\pi}{4}$, where $n$ is an integer.

Here, $x = 3\theta$. So, $3\theta = n\pi + \frac{3\pi}{4}$.

Dividing by 3, we get $\theta = \frac{n\pi}{3} + \frac{\pi}{4}$.

The principal solutions are the solutions in the interval $[0, 2\pi)$.

We find the values of $\theta$ for different integer values of $n$:

For $n=0$: $\theta = \frac{0\pi}{3} + \frac{\pi}{4} = \frac{\pi}{4}$.

For $n=1$: $\theta = \frac{\pi}{3} + \frac{\pi}{4} = \frac{4\pi + 3\pi}{12} = \frac{7\pi}{12}$.

For $n=2$: $\theta = \frac{2\pi}{3} + \frac{\pi}{4} = \frac{8\pi + 3\pi}{12} = \frac{11\pi}{12}$.

For $n=3$: $\theta = \frac{3\pi}{3} + \frac{\pi}{4} = \pi + \frac{\pi}{4} = \frac{5\pi}{4}$.

For $n=4$: $\theta = \frac{4\pi}{3} + \frac{\pi}{4} = \frac{16\pi + 3\pi}{12} = \frac{19\pi}{12}$.

For $n=5$: $\theta = \frac{5\pi}{3} + \frac{\pi}{4} = \frac{20\pi + 3\pi}{12} = \frac{23\pi}{12}$.

For $n=6$: $\theta = \frac{6\pi}{3} + \frac{\pi}{4} = 2\pi + \frac{\pi}{4} > 2\pi$.

For $n=-1$: $\theta = \frac{-1\pi}{3} + \frac{\pi}{4} = \frac{-4\pi + 3\pi}{12} = -\frac{\pi}{12} < 0$.

The principal solutions are $\left\{\frac{\pi}{4}, \frac{7\pi}{12}, \frac{11\pi}{12}, \frac{5\pi}{4}, \frac{19\pi}{12}, \frac{23\pi}{12}\right\}$.