Question

Solve: 2cos^2x – 7cosx + 3 = 0

So far I have---
2cos^2x – 7cosx = -3

but thats as far as I got :3

Answer 1

let cosx = p

Answer 2

2cos^2x – 7cosx + 3 = 0

2p^2 - 7p + 3 = 0

Answer 3

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

Answer 4

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

Answer 5

woah O.o true

Answer 6

why, why why is it always quadratics lol

Answer 7

because they're the essence of life.

Answer 8

^ OMG :')

Answer 9

I thought the essence of life would be matter

Answer 10

@isuckatmath9999 Love is not matter :-P

Answer 11

Love is relative xD

Answer 12

\[y= \cos (x)\]\[\large 2y^2 - 7y +3 =0\]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

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

cos (x)= :3

Answer 19

why is the second impossible?

Answer 20

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