Solve: 2cos^2x – 7cosx + 3 = 0
So far I have---
2cos^2x – 7cosx = -3
but thats as far as I got :3

- anonymous

Solve: 2cos^2x – 7cosx + 3 = 0
So far I have---
2cos^2x – 7cosx = -3
but thats as far as I got :3

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- ganeshie8

let cosx = p

- ganeshie8

2cos^2x – 7cosx + 3 = 0
2p^2 - 7p + 3 = 0

- ganeshie8

its a quadratic now, easily factorable... give it a try

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## More answers

- Jhannybean

do the same thing you do when you're solving a quadratic,i'll show you my method.

- anonymous

woah O.o true

- anonymous

why, why why is it always quadratics lol

- Jhannybean

because they're the essence of life.

- ParthKohli

^ OMG :')

- anonymous

I thought the essence of life would be matter

- ParthKohli

@isuckatmath9999 Love is not matter :-P

- anonymous

Love is relative xD

- Jhannybean

\[y= \cos (x)\]\[\large 2y^2 - 7y +3 =0\]multiply the leading coefficient to the constant at the end. \[\large y^2-7y +6= 0\]what two numbers give you -7 when added and 6 when multiplied? -6 and -1 \[\large (y-6)(y-1)=0\] divide by the leading coefficient you used to multiply to the end. \[\large (y-\frac{6}{2})(y-\frac{1}{2})=0 \] simplify whatever you can. \[\large (y-3)(2y-1) =0\]now resub the cos (x) back in \[\large (\cos(x)-3)(2\cos(x)-1)=0\] solve

- anonymous

I'm trying to figure out what the darned angles are

- ParthKohli

Didn't need to do that :-P\[2p^2 - 6p - p +3 = 0\]\[2p(p - 3) - 1(p - 3) = 0 \]\[(2p - 1)(p-3)=0\]

- anonymous

I appreciate the help though, this is so useful. :3

- Jhannybean

well i love that method, lol. it goes fast for me since i'm always using it to solve complex quadratics.

- ParthKohli

So in a way you just need to solve \(\cos(x) = \frac{1}{2}\) and \(\cos(x) = 3\). The second is impossibru

- Jhannybean

cos (x)= :3

- anonymous

why is the second impossible?

- ParthKohli

Because \(-1 \le \cos(x) \le 1\)

- ganeshie8

|dw:1370416709254:dw|

- Jhannybean

ganeshie had awesome mouse skills....

- ganeshie8

it never quite gets past 1

- ParthKohli

So just solve \(\cos(x) = 0.5\)

- ParthKohli

in the interval you're given, of course.

- anonymous

oh interesting

- anonymous

so the angles are 60 ... and?

- ParthKohli

Have you heard of the unit circle?

- anonymous

where the radius is always 1?

- ParthKohli

Yeah. The \(y\) coordinate is cosine

- anonymous

??

- ParthKohli

hmm

- Jhannybean

draw it outt :P

- ParthKohli

How should I explain this...|dw:1370417354072:dw|

- ParthKohli

Bad drawing, but you get the point. Or do you?

- anonymous

90,120,210&330? or am i really off

- anonymous

i know its suppposed to be based off of like Cast and stuff =/

- Jhannybean

based off whaaat?....

- ParthKohli

o_O

- ParthKohli

The second angle is clearly in the 2nd quadrant

- anonymous

my brain is really... dead

- ParthKohli

Mine too :P

- anonymous

so how in the world do i find them?

- ParthKohli

Try using WolframAlpha :P

- anonymous

i tried taht lol

- anonymous

that*

- ParthKohli

Found another solution: 300 degrees

- ParthKohli

Seems like those are the two solutions. Oh, and I messed up as always... cos(x) is the x coordinate

- ParthKohli

So yes, 60 and 300 are solutions

- anonymous

okay: so in that case: with cosx=sqrt(3)/2
x = 30° , 120°, 330°
right?

- ganeshie8

oly, x and 360-x wil give posive sqrt(3)/2
knowing x = 30, other solution wud be 360-30

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