Solve \cos(x)-\sin3x=\cos2x I have tried to rewrite as follows: \cos(2x)-\cos(x)=-\sin3x =-2\sin(\frac{3}{2}x)\sin x=-3\sin x+4\sin^3x Is

Caleb Snyder

Caleb Snyder

Answered question

2022-01-28

Solve cos(x)sin3x=cos2x
I have tried to rewrite as follows:
cos(2x)cos(x)=sin3x
=2sin(32x)sinx=3sinx+4sin3x
Is it correct? Is it possible to proceed from here or I need to transform it differently?
Transform cos2x into 12sin2x is not useful here either.

Answer & Explanation

Larissa Hogan

Larissa Hogan

Beginner2022-01-29Added 10 answers

You can rewrite it as
eix+eix2e3ixe3ix2i=e2ix+e2ix2
that becomes
e3ixe3ix+ie2ix+ie2ixieixieix=0
or, setting z=eix
z61+iz5iz2iz4+iz=0
This can be rewritten as
(z31)(z3+1)+iz2(z31)iz(z31)=0
This yields z31=0 or
z3+iz2iz+1=0
The last one can be factored as (z+i)(z2i)=0. Thus we get:
from z3=1,  e3ix=e0+2kiπ
from z+i=0,  eix=eiπ2+2kiπ
from z2i,  e2ix=eiπ2+2kiπ
Finally:
x=2kπ3
x=π2+2kπ
x=π4+kπ

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