Question

2cos^2x-cosx-1=0

Answer 1

You can think of this as simply \[2a^2-a-1=0\]where \(a = \cos x\). Solve for the value(s) of \(a\), then find the value(s) of \(\cos x\) that equal \(a\).

Answer 2

use inverse functions where needed to find all solutions of the given equation on the interval [0,2pi).

I don't understand.

Answer 3

Can you solve the equation I propose?

Answer 4

\[2a^2-a-1=0\]\[a=\]

Answer 5

\[2\cos^2x-cosx-1=0\]

Answer 6

I know...but just try it my way, okay?

Answer 7

okay

Answer 8

a=1/2

Answer 9

Mmm...are you sure?

\[2(\frac{1}{2})^2 - (\frac{1}{2}) - 1 = 0\]
\[\frac{2}{4}-\frac{1}{2} - 1 = 0\]\[\frac{1}{2}-\frac{1}{2}-1=0\]\[0-1=0\]
Oops

Answer 10

so it has to equal 0 when I plug it in?

Answer 11

Yeah! Or it isn't actually a solution...

Answer 12

So you're trying to find the values of \(a\) that make \(2a^2-a-1=0\). When you have them, you'll then find the values of \(x\) that make \(\cos x = a\). That's the part where you use the inverse functions.

Answer 13

How did you come up with 1/2 as an answer, btw?

Answer 14

2a^2-a-1=0
2a^2-a=1
a^2-a=1/2
a=1/2

Answer 15

\[2a^2-a=1\]Divide both sides by 2 and you get
\[\frac{2a^2-a}{2} = \frac{1}{2}\]\[\frac{2a^2}{2} - \frac{a}{2} = \frac{1}{2}\]

Answer 16

\[2a^2-a-1=0\]\[2a^2-2a+a-1=0\]\[(2a^2-2a)+(a-1) = 0\]\[2a(a-1)+1(a-1) = 0\]\[(2a+1)(a-1) = 0\]can you finish it from there?

Answer 17

not to sound dumb right now, but this doesn't make any sense... I don't understand how you got such a big equation

Answer 18

This is just factoring by grouping.

If you prefer, we could use the quadratic formula to solve it.

\[2x^2-x-1=0\]\[a=2,b=-1,c=-1\]\[x = \frac{-b\pm\sqrt{b^2-4ac}}{2a} = \frac{-(-1)\pm\sqrt{(-1)^2-4(2)(-1)}}{2(2)} = \frac{1\pm\sqrt{1+8}}{4}=\]

Answer 19

Or we could complete the square:

\[2a^2-a-1=0\]\[a^2-\frac{1}{2}a - \frac{1}{2} = 0\](after dividing by 2)
take half of the coefficient of \(x\), square it, and add to both sides:
\[a^2-\frac{1}{2}a - \frac{1}{2} + (\frac{1}{2*2})^2 = (\frac{1}{2*2})^2\]\[(a^2-\frac{1}{2}a + \frac{1}{16} )- \frac{1}{2} = \frac{1}{16}\]rewrite quantity in parentheses as perfect square
\[(a-\frac{1}{4})^2 -\frac{1}{2} = \frac{1}{16}\]\[(a-\frac{1}{4})^2 = \frac{9}{16}\]take square root of each side
\[a-\frac{1}{4} = \pm\frac{3}{4}\]
\[a = \frac{1\pm 3}4\]\[a = 1,~a=-\frac{1}{2}\]

Answer 20

I'll just copy somebody who did their homework tomorrow. Thank you for trying to help me understand this :)

Answer 21

testing the solutions:

\[2(1)^2-(1)-1= 0\]\[2-1-1=0\]

\[2(-\frac{1}{2})^2-(-\frac{1}{2})-1=0\]\[\frac{2}{4}+\frac{1}{2} -1 =0\]\[1-1=0\]

Answer 22

So, your job is to find the values of \(x\) such that \[\cos x = -\frac{1}2\]and \[\cos x = 1\]

Answer 23

all these slashes and parenthesis are confusing. I don't know what's up

Answer 24

all the slashes and parentheses? have you not seen fractions before?

Answer 25

this is like some beautiful mind stuff.

Answer 26

No, this is basic algebra, and I fear you are not going to have a pleasant time in trigonometry and analytic geometry if this is causing you difficulty :-(

Answer 27

im dyslexic

Answer 28

That's not going make things easier, I agree!

Answer 29

and i have ADHD... life is a struggle