\(\displaystyle{\sin{{\left({4}{x}\right)}}}=-{2}{\sin{{\left({2}{x}\right)}}}\) solutions in \(\displaystyle{\left[{0},{2}\pi\right)}\)

juctommaccedo662f

juctommaccedo662f

Answered question

2022-03-30

sin(4x)=2sin(2x) solutions in [0,2π)

Answer & Explanation

Drake Huang

Drake Huang

Beginner2022-03-31Added 15 answers

Use the following identities
sin(2x)=2sin(x)cos(x)
cos(2x)=2cos2(x)1
sin(4x)=sin(2x)cos(2x)=8sin(x)cos3(x)4sin(x)cos(x)
which simplifies to
8sin(x)cos3(x)=0
with solution x=kπ2 since either sin(x) has to be zero or cos(x) has to be zero.

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?