Linearization formulas:

[Aladdin]

Solve the equation:
cos(x)+cos(3x)+cos(5x)=2cos(2x)+1

What i though about is the following:
I took: cos(6x/2)*cos(-4x/2)+cos(3x)=2cos(2x)+1(( IM NOT SURE IF I LEAVE IT LIKE THAT THE 2COS(2X)+1 ???))
then: common factor: cos(3x){(cos(2x)+1)}=2cos(2x)+1
And then I have no idea.

Thank you in advance. --------------- I will appreciate any help----------

 

[Denis]

Too much typing for me!!
Go here to get an idea:
http://answers.yahoo.com/question/index ... 631AAjyeMF

 

[galactus]

We can try rewriting it and using a u sub, then factoring.

Rewrite as:

\(\displaystyle 3cos(x)-16sin^{2}(x)cos^{3}(x)=4cos^{2}(x)-1\)

\(\displaystyle 3cos(x)-16(1-cos^{2}(x))cos^{3}(x)=4cos^{2}(x)-1\)

\(\displaystyle 3cos(x)-16cos^{3}(x)-4cos^{2}(x)+16cos^{5}(x)+1=0\)

Let \(\displaystyle u=cos(x)\)

\(\displaystyle 16u^{5}-16u^{3}-4u^{2}+3u+1=0\)

Factor:

\(\displaystyle (u-1)(2u-1)(2u+1)^{3}=0\)

Now finish?.

 

[Aladdin]

aha, thank you Mr galactus .,. i appreciate your precious time,