Question

Find all solutions in the interval [0, 2π).

cos^2x + 2 cos x + 1 = 0

Answer 1

@amistre64

Answer 2

we have a quadratic equation ....

Answer 3

how do we solve a quadratic equation?

Answer 4

if we let u = cos(x)

\[cos^2x + 2 ~cos x + 1 = 0\]
\[u^2 + 2u + 1 = 0\]

Answer 5

\[u^2+2u+1=0\]
\[(1+u) (1+u)\]

Answer 6

@amistre64 where do I go from here?

Answer 7

hmmmmm

Answer 8

If (u+1)^2 or (u+1)*(u+1) = 0, what solutions do you have for u ?

What can you tell about at least one of those two numbers if you know that when they are multiplied with each other they equal 0 ?

What can you tell about A (or B) if you know that A * B = 0 ?

Answer 9

This is kinda hard.....

Answer 10

WHat grade are you in @dglicks

Answer 11

@AngusV if i know that A*B=0 then one of those variables is 0

Answer 12

Perfect. In this case both variables are equal to each other ( so it's kinda like A*A=0 ). Which means A is 0.

Now instead of A you have (u+1). Which means (u+1)=0. Which means u=(-1). Recall that u was just a notation (our way of saying) cos(x). That means cos(x)=-1.

Answer 13

got it thank you @AngusV

Answer 14

x = 2π
x = π
x = \[\pi/4, 7 \pi/ 4\]
x = \[\pi/2 , 3\pi/2\]
@AngusV these are my options

Answer 15

The only "place" on the trigonometric circle where cos(x)= -1 is on the far (furthest) left - at x=180 degrees or x=pi. At pi/2 and 3pi/2 cos(x) = 0 but you need cos(x) to equal -1. Where did you get those values from ?

Answer 16

Those values are my options for the multiple choice question

Answer 17

@AngusV

Answer 18

@campbell_st

Answer 19

so its the orginal question

\[\cos^2(x)+2\cos(x) + 1 = 0\]

Answer 20

@dglicks is this the question?

Answer 21

No the question is:
Find all solutions in the interval: \[(0,2\pi)\]
\[\cos^2x + 2 \cos x + 1 = 0 \]

Answer 22

So yes that is PART of the question

Answer 23

ok... so this is a quadratic equation that can be factored... its a perfect square

you used the substitution method and came up with

\[(1 + u)(1+u) = (u + 1)^2\]

if you reverse the subsitution you get

\[(\cos(x) + 1)^2 = 0\]

so now you need to solve

cos(x) + 1 = 0 for cos(x)

can you do that..?

Answer 24

Im not exactly sure what to do in order to solve for cos(x) + 1= 0

Answer 25

well get cos(x) on its own by subtracting 1 from both sides of the equation

Answer 26

cos(x)= -1
I know how to do that, and to solve for this do I simply put that in my calculator?

Answer 27

well remember you are working in radians

you can also use the cos curve