Solving \sin 2x \sin x+\cos^2 x=\sin 5x \sin 4x+\cos^2 4x

Rudy Koch

Rudy Koch

Answered question

2022-01-23

Solving sin2xsinx+cos2x=sin5xsin4x+cos24x

Answer & Explanation

oferenteoo

oferenteoo

Beginner2022-01-24Added 12 answers

Use cos2(a)+sin2(a)=1 first on both sides then factorize.
sin(x)[sin(2x)sin(x)]=sin(4x)[sin(5x)sin(4x)]
Now use sin(A)sin(B) as a product on both sides and simplify (giving some roots sin(x2)=0)
sin(x)2cos(3x2)sin(x2)=sin(4x)2cos(9x2)sin(x2)
then write sin(C)cos(D) as a difference of sines on both sides and simplify.
12sin(5x2)+sin(x2)=12sin(17x2)+sin(x2)
Write the difference of sines as a product and solve
sin(5x2)sin(17x2)=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?