Question

Find all solutions to the equation.

cos2x + 2 cos x + 1 = 0

Answer 1

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

Answer 2

I knew you had to use sine but I wasn't sure how or why.. still a little confused

Answer 3

Try and substitute with those two, simplify and tell me what you get.

Answer 4

okayy hold on

Answer 5

The over all goal is to simplify as much as possible so we get an easy expression to solve for x.

Answer 6

I got a weird simplification for the first equation I don't think I'm doing it right

Answer 7

And some times you have to substitute with trig-identities to achieve that.

Answer 8

Okay write it here and we'll see

Answer 9

Just the expression after substitution without any simplification.

Answer 10

I feel like it's too far off to even write. I'm not completely sure how I'm supposed to simplify the terms

Answer 11

cos(2x) + 2 cos(x) + 1 = 0 \[\cos^2(x) - \sin^2(x) + 2 \cos(x) + \cos^2(x) + \sin^2(x) = 0\]The sines cancel and you get \[2\cos^2(x) + 2 \cos(x) = 0\]\[\cos (x) (\cos (x) + 1) = 0\]

Answer 12

From there you'll find all solutions when
cos(x) = 0 or cos(x) + 1 = 0

Answer 13

So which parts do I need to solve to determine if it equals 0 for cos(x) or cos(x)+1?

Answer 14

x = arccos(0) or x = arccos(-1).

Answer 15

Dont forget that both cos(90 deg) and cos(270 deg) is 0.

Answer 16

arccos(0)= pi/2
arccos(-1)= pi

Answer 17

So then I plug those into cos(x)=0 and cos(x)+1=0 ?

Answer 18

That's only necessary if you want to check if the answer is cocrrect. The assignment is to find the value of x, right?

And one soulution is
x = arccos(0)
=> x = 90 deg = pi/2
x = 270 deg = 3pi/2

x = arccos(-1)
=> x = 180 deg = pi

Answer 19

So there are a total of three solutions which are pi/2, 3pi/2, and pi?

Answer 20

No, there are actually an infinte amount of soulutions and that's one possible soulution.

Answer 21

Finding the other ones are simple though.

Answer 22

cos is periodic.
-1 = cos(pi) = cos(3pi) = cos(9pi) and so on
so all you have to do is to add a term for each partial soulution to cover for the periodicity.

x = arccos(0)
=> x = pi n - pi/2 for any whole number n

x = arccos(-1)
=> x = 2n pi - pi for any whole number n
x = 2k pi + pi for any whole number k