The Problem: Solve the equation: \(\displaystyle{\cos{{x}}}={\cos{{3}}}{x}+{2}{\sin{{2}}}{x}\) The Result: \(\displaystyle{x}={k}{\frac{{\pi}}{{{2}}}},{k}\in{\mathbb{{{Z}}}}\) My

Brielle James

Brielle James

Answered question

2022-04-11

The Problem: Solve the equation:
cosx=cos3x+2sin2x
The Result:
x=kπ2,kZ
My solution:
cosx=4cos3x-3cosx+4sinxcosx
4cos3x-4cosx+4sinxcosx=0
4cosx(cos2x+sinx-1)=0
cosx(1-sin2x+sinx-1)=0
cosx(sinx-sin2x)=0
cosxsinx(1-sinx)=0
x=π2+lπx=mπx=π2+2nπ; l,m,n
Question: Is my solution equal to the result?

Answer & Explanation

Terzago66cl

Terzago66cl

Beginner2022-04-12Added 8 answers

My resolution:
cosx=cos(3x)+2sin(2x)
cos(3x)2sin(2x)+cos(x)=0
Using the following identity
cos(p)+cos(q)=2sin(p+q2)sin(pq2)
I obtaining,
2sin(2x)+2sin(x+3x2)sin(x+3x2)=0
Hence
2sin(2x)+2sin(2x)sin(x)=0
Factorizing
2sin(2x)(sin(x)1)=0
Solving each part separately:
sin(2x)=0orsin(x)1=0
I have your solutions:
sin(2x)=0:x=kπ,:x=π2+kπ
sin(x)1=0:x=π2+2kπ,  kZ

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