anonymous
  • anonymous
Solve: 2cos^2x – 7cosx + 3 = 0 So far I have--- 2cos^2x – 7cosx = -3 but thats as far as I got :3
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
ganeshie8
  • ganeshie8
let cosx = p
ganeshie8
  • ganeshie8
2cos^2x – 7cosx + 3 = 0 2p^2 - 7p + 3 = 0
ganeshie8
  • ganeshie8
its a quadratic now, easily factorable... give it a try

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More answers

Jhannybean
  • Jhannybean
do the same thing you do when you're solving a quadratic,i'll show you my method.
anonymous
  • anonymous
woah O.o true
anonymous
  • anonymous
why, why why is it always quadratics lol
Jhannybean
  • Jhannybean
because they're the essence of life.
ParthKohli
  • ParthKohli
^ OMG :')
anonymous
  • anonymous
I thought the essence of life would be matter
ParthKohli
  • ParthKohli
@isuckatmath9999 Love is not matter :-P
anonymous
  • anonymous
Love is relative xD
Jhannybean
  • 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
  • anonymous
I'm trying to figure out what the darned angles are
ParthKohli
  • 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
  • anonymous
I appreciate the help though, this is so useful. :3
Jhannybean
  • Jhannybean
well i love that method, lol. it goes fast for me since i'm always using it to solve complex quadratics.
ParthKohli
  • ParthKohli
So in a way you just need to solve \(\cos(x) = \frac{1}{2}\) and \(\cos(x) = 3\). The second is impossibru
Jhannybean
  • Jhannybean
cos (x)= :3
anonymous
  • anonymous
why is the second impossible?
ParthKohli
  • ParthKohli
Because \(-1 \le \cos(x) \le 1\)
ganeshie8
  • ganeshie8
|dw:1370416709254:dw|
Jhannybean
  • Jhannybean
ganeshie had awesome mouse skills....
ganeshie8
  • ganeshie8
it never quite gets past 1
ParthKohli
  • ParthKohli
So just solve \(\cos(x) = 0.5\)
ParthKohli
  • ParthKohli
in the interval you're given, of course.
anonymous
  • anonymous
oh interesting
anonymous
  • anonymous
so the angles are 60 ... and?
ParthKohli
  • ParthKohli
Have you heard of the unit circle?
anonymous
  • anonymous
where the radius is always 1?
ParthKohli
  • ParthKohli
Yeah. The \(y\) coordinate is cosine
anonymous
  • anonymous
??
ParthKohli
  • ParthKohli
hmm
Jhannybean
  • Jhannybean
draw it outt :P
ParthKohli
  • ParthKohli
How should I explain this...|dw:1370417354072:dw|
ParthKohli
  • ParthKohli
Bad drawing, but you get the point. Or do you?
anonymous
  • anonymous
90,120,210&330? or am i really off
anonymous
  • anonymous
i know its suppposed to be based off of like Cast and stuff =/
Jhannybean
  • Jhannybean
based off whaaat?....
ParthKohli
  • ParthKohli
o_O
ParthKohli
  • ParthKohli
The second angle is clearly in the 2nd quadrant
anonymous
  • anonymous
my brain is really... dead
ParthKohli
  • ParthKohli
Mine too :P
anonymous
  • anonymous
so how in the world do i find them?
ParthKohli
  • ParthKohli
Try using WolframAlpha :P
anonymous
  • anonymous
i tried taht lol
anonymous
  • anonymous
that*
ParthKohli
  • ParthKohli
Found another solution: 300 degrees
ParthKohli
  • ParthKohli
Seems like those are the two solutions. Oh, and I messed up as always... cos(x) is the x coordinate
ParthKohli
  • ParthKohli
So yes, 60 and 300 are solutions
anonymous
  • anonymous
okay: so in that case: with cosx=sqrt(3)/2 x = 30° , 120°, 330° right?
ganeshie8
  • 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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