Trig Help...

[jacob0401]

cos 2x = 2 cos x

I'm supposed to find all the exact solutions(2 of them), algebraically, in radians.

This has to do with double angle identities, but I can't get it to a point where I can factor.

Any help is appreciated...

btw... cos 2x = 1-2sin^2(x) or 2cos^2(x)-1 or cos^2(x)-sin^2(x)

 

[dighn]

nm... cos2x is 2cos^2x-1 not cos^2x -1

 

[jacob0401]

Yeah I can get it down to quadratic, but that gives me a decimal or root answer and it specifies radian measure.

 

[dighn]

.

 

[UncleWai]

memorize the unit circle, calculator is evil for trig.

 

[jacob0401]

Originally posted by: UncleWai
memorize the unit circle, calculator is evil for trig.

i have memorized it...but i can't even reduce it to the point where i can solve for cos/sin/tan...

 

[aux]

If y = cos(x); you get:
2y^2 - 1 = y, or equivalently:
2y^2 - y - 1 = 0, which gives
y = 1 or y = -1/2

 

[simms]

i hate trig.

 

[pennylane]

trig calms my soul.

 

[UncleWai]

I posted when dighn had an "answer"

you wrote£¼"but that gives me a decimal or root answer and it specifies radian measure. "

So I thought you were using a calculator to find what X IS 😱

 

[Syringer]

Originally posted by: aux
If y = cos(x); you get:
2y^2 - 1 = y, or equivalently:
2y^2 - y - 1 = 0, which gives
y = 1 or y = -1/2

It's actually 2y^2 - 1 = 2y

 

[Syringer]

And when solving that I get y = +-(sqrt(3)+1)/2..see if you can convert that to polar after taking cos^-1 of that..

Or you might just be able to leave it as cos^-1((sqrt(3)+1)/2)

 

[aux]

Originally posted by: Syringer
And when solving that I get y = +-(sqrt(3)+1)/2..see if you can convert that to polar after taking cos^-1 of that..

Or you might just be able to leave it as cos^-1((sqrt(3)+1)/2)

yes, it's 2y; doing math after midnight is never a good thing 🙂

btw, cos^(-1) of ((sqrt(3)+1)/2) does not exist because ((sqrt(3)+1)/2) > 1 (this is because cos takes values in [-1,1] only)

the answer should be cos^(-1) of ((-sqrt(3)+1)/2) (there are more solutions using the periodicity of cos, and also cos(-y) = cos y; probably you will need to find the answers in [0,2*pi) only, because of the periodicity)

check the place of the - sign again

edit: typo, it's too early for me to think

 

[speg]

Special Triangles?