How do I solve this equation: \(\displaystyle{\sin{{x}}}-{3}{\sin{{2}}}{x}+{\sin{{3}}}{x}={\cos{{x}}}-{3}{\cos{{2}}}{x}+{\cos{{3}}}{x}\)

Delilah Novak

Delilah Novak

Answered question

2022-04-18

How do I solve this equation:
sinx3sin2x+sin3x=cosx3cos2x+cos3x

Answer & Explanation

tutaonana223a

tutaonana223a

Beginner2022-04-19Added 15 answers

First
sin(x)+sin(y)=2sin(x+y2)cos(xy2)
cos(x)+cos(y)=2cos(x+y2)cos(xy2)
Therefore
sin(x)+sin(3x)=2sin(2x)cos(x), cos(x)+cos(3x)=2cos(2x)cos(x)
Hence, we can write
2sin(2x)cos(x)3sin(2x)=2cos(2x)cos(x)3cos(2x)
Factoring out we get
sin(2x)[2cos(x)3]=cos(2x)[2cos(x)3]
Eventually sin(2x)=cos(2x) Or cos(x)3=0

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?