Simplifying a trigonometric expression

[Petkovsky]

cos(4x) + $$\sqrt{2}$$cos(3x) + cos(2x) = 0
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Ok, so break down everything to a double angle and i get:

2cos^2(2x) - 1 + sqrt(2)*(sqrt((cos(2x) + cos^2(2x))/2) - sqrt((1-cos^(2x)*(1-cos(2x))/2) + cos(2x) = 0

...which is quite complicated to solve even if i substitute cos(2x) with 't'.

I also tried another aproach:

2cos^2(2x) - 1 + sqrt(2)*(cos(2x)cos(x) - sin(2x)(sin(x)) + 2cos(2x) - 1 = 0

here i aimed to complete a square but i can't see how.

Can you please give me some advice on how to continue or use another method

 

[konthelion]

Hello.

I used the Sum and Difference Formula for cosines.
Def: cos(a)+cos(b) = 2*cos((a+b)/2)*cos((a-b)/2)

Then, I let a=4x and b=2x

The original equation's LHS becomes
= 2cos((4x+2x)/2)*cos(4x-2x/2)) + $$\sqrt{2}$$cos(3x)
= 2cos(3x)cos(2x) + $$\sqrt{2}$$cos(3x) by simplifying
= cos(3x)[2cos(2x) + $$\sqrt{2}$$] by distributive property
Thus, cos(3x) = 0 or 2cos(2x) + $$\sqrt{2}$$ = 0

Hope this helped.

 

[Petkovsky]

Thank you

 

[Petkovsky]

Hello.

= 2cos((4x+2x)/2)*cos(4x-2x/2)) + $$\sqrt{2}$$cos(3x)
= 2cos(3x)cos(2x) + $$\sqrt{2}$$cos(3x) << -- didnt you make a mistake here?

Didnt you make a mistake?
Shouldn't it be = 2cos(3x)cos(x) + $$\sqrt{2}$$cos(3x)?

Anyway thanks for the help.

 

[konthelion]

Didnt you make a mistake?
Shouldn't it be = 2cos(3x)cos(x) + $$\sqrt{2}$$cos(3x)?

Anyway thanks for the help.

Yes, sorry.

 

[Petkovsky]

sin(3x) = cos(x-30)
How about this one?

Where should I start from?

 

[Petkovsky]

Ok i got it :)