quick prob (1 Viewer)

[googleplex]

general solution of

cos4x + cos2x = sqrt2 . cos^2(x) + 1/sqrt2 . sin2x

 

[Slidey]

DISCLAIMER: I am not responsible for brain damage caused by reading this post.

cos4x=cos^2(2x)-sin^2(2x)=cos2x.cos2x-sin2x.sin2x=cos^4(x) - 2cos^2(x)sin^2(x) + sin^4(x) - 4cos^2(x)sin^2(x)
cos2x=cos^2(x)-sin^2(x)
sin2x=2cosx.sinx
so...

LHS = cos^4(x) - 2cos^2(x)sin^2(x) + sin^4(x) - 4cos^2(x)sin^2(x) + cos^2(x)-sin^2(x)
cosx=c
sinx=s

(c+sqrt(3+2sqrt(2))s)(c-sqrt(3+2sqrt(2))s)(c+sqrt(3-2sqrt(2))s)(c-sqrt(3-2sqrt(2))s)=((sqrt(2)-1)c+s)(c+s)

(cosx+sqrt(3+2sqrt(2))sinx)(cosx-sqrt(3+2sqrt(2))sinx)(cosx+sqrt(3-2sqrt(2))sinx)(cosx-sqrt(3-2sqrt(2))sinx)=((sqrt(2)-1)cosx+sinx)(cosx+sinx)

My calculator crashed and erased all it's files when I put your question into it to check my answer.

But it was fun, anyway.

 

[mojako]

See the attachment.
I did it by inspection


cos4x + cos2x = sqrt2 . cos^2(x) + 1/sqrt2 . sin2x
2cos^2 2x - 1 + cos 2x = sqrt(2) (1/2 + 1/2 cos 2x) + 1/sqrt(2) sin 2x
2cos^2 2x - 1 = sqrt(2) (1/2 + 1/2 cos 2x) + [1/sqrt(2) sin 2x - cos 2x]
transform the [...] into one cos function using auxiliary/subsidiary angle.
solve the quadratic in cos 2x.

alternatively, use the t-formulae instead of doing the transformation.

Oh by the way, there's something hidden on your screen, can you see it?
if it's hidden then it can't be seen... what a stupid question.

 

[McLake]

Arg, by brain exploded and my eyes burst.

Damn you Slide Rule.

 

[d_elmo]

whats sqrt?

 

[Supra]

square root

 

[Jase]

to this day, i dont know what general solution is.

 

[ngai]

general solution of

cos4x + cos2x = sqrt2 . cos^2(x) + 1/sqrt2 . sin2x

by inspection:
x = -1/2*Pi, 1/2*Pi, -1/8*Pi, 7/8*Pi, arctan(2^(1/2)*(2+(2+2^(1/2))^(1/2))-3*2^(1/2)-1), arctan(2^(1/2)*(2+(2+2^(1/2))^(1/2))-3*2^(1/2)-1)-Pi, -arctan(-2^(1/2)*(2-(2+2^(1/2))^(1/2))+3*2^(1/2)+1), -arctan(-2^(1/2)*(2-(2+2^(1/2))^(1/2))+3*2^(1/2)+1)+Pi

 

[mojako]

ngai did that while he was eating pancake with maple syrup, which is the secret to success in the HSC


general solution... solution which is general enough to be understood by the ordinary people of our society, i.e. doesn't contain mathematical jargons

 

[Xayma]

by inspection:
x = -π/2, π/2, -π/8, 7π/8, arctan(2^1/2*(2+(2+2^1/2)^1/2)-3*2^1/2-1), arctan(2^1/2*(2+(2+2^1/2)^1/2)-3*2^1/2-1)-π, -arctan(-2^1/2*(2-(2+2^1/2)^1/2)+3*2^1/2+1), -arctan(-2^1/2*(2-(2+2^1/2)^1/2)+3*2^1/2+1)+π

What about 5π/2? Or should I just assume to add 2nπ to each.

 

[mojako]

What about 5π/2? Or should I just assume to add 2nπ to each.

hmm.. apparently maple syrup isnt that useful...

oh thanks for the π symbol.

 

[Xayma]

Well (s)he had it but forgot to make it general by adding 2nπ to each one.
Eww that π symbol looks crap put it in Times New Roman =P

 

[mojako]

I believe it's a "he".
Do some inspection!!
inspection ~ investigation,
where ~ is approximately equal to.

Oh I like Arial coz its quicker to write
If I dont put anything it looks like an "n".
apparently you can just write "Times" (with no quotes its okay)