Solving sin⁡3θ=12 on the interval [0,2π]. I don't understand where some solutions came from.

Alvin Pugh

Alvin Pugh

Answered

2022-01-25

Solving sin3θ=12 on the interval [0,2π]. I don't understand where some solutions came from.

Answer & Explanation

Frauental91

Frauental91

Expert

2022-01-26Added 15 answers

Write out 3θ:π6,5π6,13π6,17π6,25π6,29π6. Now subtract π6 from 13π6, and you get 12π6=2π. The sine function is periodic, with this period. You get all the other by adding 2π  or  4π to the first solutions.
Maybe a more obvious way of thinking about the problem is to say α=3θ, and find all solutions of sinα=12 in the interval [0,6π]

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