Linearization formulas:

Aladdin

Full Member
Joined
Mar 27, 2009
Messages
553
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----------
 
We can try rewriting it and using a u sub, then factoring.

Rewrite as:

3cos(x)16sin2(x)cos3(x)=4cos2(x)1\displaystyle 3cos(x)-16sin^{2}(x)cos^{3}(x)=4cos^{2}(x)-1

3cos(x)16(1cos2(x))cos3(x)=4cos2(x)1\displaystyle 3cos(x)-16(1-cos^{2}(x))cos^{3}(x)=4cos^{2}(x)-1

3cos(x)16cos3(x)4cos2(x)+16cos5(x)+1=0\displaystyle 3cos(x)-16cos^{3}(x)-4cos^{2}(x)+16cos^{5}(x)+1=0

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

16u516u34u2+3u+1=0\displaystyle 16u^{5}-16u^{3}-4u^{2}+3u+1=0

Factor:

(u1)(2u1)(2u+1)3=0\displaystyle (u-1)(2u-1)(2u+1)^{3}=0

Now finish?.
 
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