Alvin Pugh

Answered

2022-01-25

Solving $\mathrm{sin}3\theta =\frac{12}{}$ on the interval $\left[0,2\pi \right]$. I don't understand where some solutions came from.

Answer & Explanation

Frauental91

Expert

2022-01-26Added 15 answers

Write out $3\theta :\frac{\pi }{6},\frac{5\pi }{6},\frac{13\pi }{6},\frac{17\pi }{6},\frac{25\pi }{6},\frac{29\pi }{6}$. Now subtract $\frac{\pi }{6}$ from $\frac{13\pi }{6}$, and you get $\frac{12\pi }{6}=2\pi$. The sine function is periodic, with this period. You get all the other by adding to the first solutions.
Maybe a more obvious way of thinking about the problem is to say $\alpha =3\theta$, and find all solutions of $\mathrm{sin}\alpha =\frac{1}{2}$ in the interval $\left[0,6\pi \right]$

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