Question

Double angle formulas:
cos 4x - cos 2x = 0

Answer 1

I dont know what to do with the cos 4x, however I understand what to do with the cos 2x

Answer 2

cos 4x - cos 2x = 0
cos 4x - (2cos^2 x - 1) = 0

Answer 3

The cos 4x is a mystery to me

Answer 4

\[\cos(2x) = 2\cos^2(x)-1 \implies \cos(4x) = 2\cos^2(2x)-1\]

Answer 5

Does that help?/

Answer 6

Oh ok. and then do I use the same identity again on 2cos^2(2x) -1?

Answer 7

somehow you're going to have to turn it into a quadratic.

Answer 8

No you just have to change \(\cos(4x)\) to turn it into a quadratic.

Answer 9

So you will replace \(\color{red}{\cos(4x)}\) with \(\color{red}{2\cos^2(2x)-1}\)\[\color{red}{2\cos^2(2x)} -\cos(2x)\color{red}{-1} = 0\]

Answer 10

I factor this now?

Answer 11

Then you can make this an easier function to solve for by: \(\color{blue}{a=\cos(2x)}\)\[2\color{blue}{a^2}-\color{blue}a-1=0\]

Answer 12

So I dont factor. I just do that?

Answer 13

Or do I factor then do that

Answer 14

No, you have to factor. I just helped making it easier to factor by substituting.

Answer 15

Factor: \(2a^2-a-1=0\)

Answer 16

(a-2)(a+1)

Answer 17

Or (2a-2)(a+1)?

Answer 18

No (2a-1)(a+1)

Answer 19

\[(2a-1)(a+1) = 2a^2 +2a-a-1 = 2a^2 +a-1 \ne 2a^2-a-1\]

Answer 20

!! (2a+1)(a-1)

Answer 21

\[(2a+1)(a-1) = 2a^2 -2a+a-1 = 2a^2 -a-1 ~\checkmark\]

Answer 22

Now that we have factored it, resubstitute the \(a\)

Answer 23

(2cos(x)+1)(cos(x)-1)

Answer 24

Then (2cos(x)+1) = 0 and (cos(x)-1) = 0

Answer 25

No, tha was not the correct substitution.

Answer 26

Remember, \(a = \cos(2x)\)

Answer 27

(2cos(2x)+1) = 0 and (cos(2x)-1)

Answer 28

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

Answer 29

Yes.

Answer 30

So now I use the double angle on both equations right?

Answer 31

(4cos^2(x)-1+1) = 0 and (2cos^2(x)-1-1)

Answer 32

So 4cos^2(x) = 0
2cos^2(x) -2 =0

Answer 33

4cos^2(x) = 0
cos^2(x) = 1

Answer 34

Hmm, trying to figure it out.

Answer 35

arccos(0) = pi/2 and 3pi/2
arccos(sqrt(1)) = 0

Answer 36

?

Answer 37

I'll check my answer with yours in a sec,

Answer 38

\[2\cos(2x) +1 = 0\]\[2\cos(2x)=-1\]\[\cos(2x)=-\frac{1}{2}\]Take the inverse to solve for \(2x\)\[2x= \cos^{-1}\left(-\frac{1}{2}\right)\]

Answer 39

What radians give -1/2?

Answer 40

7pi/6 and 11pi/6

Answer 41

And then I multiply that by 2?

Answer 42

Not exactly.

Answer 43

cosine is negative in Q2.

Answer 44

If I'm trying to find a negative x isn't cos negative?

Answer 45

and 3, but \(\frac{7\pi}{6} = -\dfrac{\sqrt{3}}{2}\)

Answer 46

OH I did sine didnt I.

Answer 47

Yes.

Answer 48

4pi/3 and 5pi/6

Answer 49

Yes and no.

Answer 50

Cos cannot be negative?