The principal solutions of tan 3 theta = -1 a | vedclass

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(A) Given $	an 3 	heta = -1$. Since $	an \frac{3 \pi}{4} = -1$,we have $	an 3 	heta = 	an \frac{3 \pi}{4}$. The general solution is $3 	heta =

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

Vedclass · Answer

(A) Given $	an 3 	heta = -1$. Since $	an \frac{3 \pi}{4} = -1$,we have $	an 3 	heta = 	an \frac{3 \pi}{4}$. The general solution is $3 	heta = n \pi + \frac{3 \pi}{4}$,where $n \in \mathbb{Z}$. Dividing by $3$,we get $	heta = \frac{n \pi}{3} + \frac{\pi}{4}$. For principal solutions,we consider $	heta \in [0, 2 \pi)$. For $n = 0$,$	heta = \frac{\pi}{4}$. For $n = 1$,$	heta = \frac{\pi}{3} + \frac{\pi}{4} = \frac{7 \pi}{12}$. For $n = 2$,$	heta = \frac{2 \pi}{3} + \frac{\pi}{4} = \frac{11 \pi}{12}$. For $n = 3$,$	heta = \pi + \frac{\pi}{4} = \frac{5 \pi}{4}$. For $n = 4$,$	heta = \frac{4 \pi}{3} + \frac{\pi}{4} = \frac{19 \pi}{12}$. For $n = 5$,$	heta = \frac{5 \pi}{3} + \frac{\pi}{4} = \frac{23 \pi}{12}$. Thus,the set of solutions is $\left\{\frac{\pi}{4}, \frac{7 \pi}{12}, \frac{11 \pi}{12}, \frac{5 \pi}{4}, \frac{19 \pi}{12}, \frac{23 \pi}{12}ight\}$.

The principal solutions of $\tan 3 \theta = -1$ are

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