Question

can someone show me the first step in how to solve the trig equation: 2cos^2x-cosx-1=0.
the restrictions are 0 12 years ago

Answer 1

im not sure what identity to use here

Answer 2

is it \[\cos^2 x \] or \[\cos ^{2x}\]?

Answer 3

cos\[\cos ^{2} x\]

Answer 4

\[2\cos ^{2}x-cosx-1\] okay. well do you know the identity for\[\cos ^{2}x\]?

Answer 5

1+cos(2x)/2 ??

Answer 6

Yes :D so now you can replace it ! \[2(\frac{ 1+\cos(2x) }{ 2 })-cosx-1=0\]

Answer 7

Oh wait, i'm sorry,you don't even have to do this. this is much like the previous question you asked. Sorry sorry.

Answer 8

haha im so confused on how to begin this because I have the same equation written down but I get totally stuck after that.

Answer 9

So lets start over. D: I jumped ahead of myself. You have \[2\cos ^{2}x-cosx-1=0\] you can write that in terms of x, right? \[2x^2-x-1=0\]

Answer 10

yeah...

Answer 11

So factoring that our you have \[2x^2-x-1=0\] and this is a quadratic equatino solving for x! :) where you'll have 2 values for x. \[x^2-x-2=(x-\frac{ 2 }{ 2 })(x+\frac{ 1 }{ 2 })\]\[(x-1)(2x+1)\]

Answer 12

substitute your \[x=\cos x\] BACK into the equation and you'll get \[(\cos x-1)(2\cos x+1)\] and now we have to solve each equation separately. \[\cos x=1\] and \[2\cos x=-1 \rightarrow \cos x=-\frac{ 1 }{ 2 }\]

Answer 13

Do you see where i'm going with this?

Answer 14

no not really I have been using a different method to solve them? haha

Answer 15

Oh.. which method have you been using to solve them?

Answer 16

Basically you're evaluating the equation to find x-values that correspond with the angle measurement. so if cos (x)=1, you're trying to find all the radian measurements where cos (x) will equal 1. and the same thing with cos (x) = -1/2

Answer 17

okay I have been doing my whole workseet wrong because I didn't understand what I was trying to do. but that makes sence but then again I might not be doing it wrong. for example, one that I thought I got the answer to is \[\cos ^{2} x-3\sin x=3\]
I worked it out like this :

\[1-\sin ^{2} x-3sinx=3\]
\[\sin ^{2} x+3 sinx+2\]
\[sinx+1(sinx+2)\]
=sin(x)=-1
-1=90degrees
so x=90degrees

Answer 18

There are many ways to work the problem, but you are working it fine :) As long as you are finding the radian measurement of the angle in question.You got it. But did you equal both of those to zero? \[sin(x) +1 =0\] and \[\sin(x)+2 =0\]?

Answer 19

whoops!(: I was thinking to far in advanced that's a mistake!

Answer 20

Mmhmm

Answer 21

Understand how to solve this problem now? :P

Answer 22

yesss haha thank you!

Answer 23

No problem! ^_^