Question

How would you solve 2cos^2(4x)-1=0

Answer 1

Do you remember your Cosine Double Angle Formula?
It shows up in 3 different forms, but this is one of them,\[\large\rm 2\cos^2\color{orangered}{x}-1=\cos(2\color{orangered}{x})\]

Answer 2

So if you're given\[\large\rm 2\cos^2\color{orangered}{4x}-1\]Do you understand how to use the formula to deal with it?

Answer 3

it's just in the form of a quadratic equation.. I have to factor it?

Answer 4

You can do that if you like...
but did you not read anything that I wrote...?

Apply the formula.

Answer 5

Our formula tells us\[\large\rm 2\cos^2\color{orangered}{x}-1=\cos(2\color{orangered}{x})\]So applying the formula to our problem,\[\large\rm 2\cos^2\color{orangered}{4x}-1=\cos(2\cdot\color{orangered}{4x})\]yes?

Answer 6

\[\large\rm 2\cos^2\color{blue}{x}-1=\cos(2\color{blue}{x})\]

Isn't it just the same exact thing..?

Answer 7

I don't know what you're asking :(

Answer 8

You're trying to cancel out the 4x?

Answer 9

No.
I was trying to turn \(\large\rm 2cos^2(4x)-1\) into something simpler like \(\large\rm cos(8x)\)

Answer 10

so you basically got rid of the 4x on the left side

Answer 11

The whole mess of stuff turned into a single cosine.
The angle changed as well.

Answer 12

We didn't really care about the 4x,
it didn't matter what the angle was.

It was the fact that this identity gets rid of the -1 and the 2, and the square.
That's why we're applying this identity.

Answer 13

What are we trying to do here, set cosine alone on one side?

Answer 14

Our problem,\[\large\rm 2\cos^2(4x)-1=0\]after applying our Cosine Double Angle Formula becomes,\[\large\rm \cos(8x)=0\]which is a lot easier to solve.
It avoids a bunch of unnecessary Algebra steps.

Answer 15

If that's too confusing though,
you can take the Algebra route instead.

Answer 16

Where did the -1 go?

Answer 17

Add 1 to both sides,\[\large\rm 2\cos^2(4x)=1\]Divide by 2,\[\large\rm \cos^2(4x)=\frac12\]Square root,\[\large\rm \cos(4x)=\pm\frac{\sqrt2}{2}\]If you prefer to do it this way, you can..
very tedious though.

Answer 18

What do you mean where did it go?

Have you learned your Double Angle Formulas yet? 0_o

Answer 19

I'll just watch a quick video about that on youtube rn brb

Answer 20

formula for \(\cos(2t)\) is \(2\cos^2t - 1\)

Answer 21

do I need to memorize these formulas?

Answer 22

I understand that he original equation turns to cos(8x) when the formula is applies. I'm afraid that I won't be able to recognize it

Answer 23

\[ \cos(8x)=0 \]\[\cos(8x \div 8 ) = 0 \div8 = 0\]

Answer 24

Woah woah woah :) Don't try to get fancy with the 8.
No bueno!

Answer 25

some website is telling me to use the zero product property and set the cos to just "a"

Answer 26

How else will we solve that problem?

Answer 27

Ignore the fact that you have an 8 in there that is making you scratch your head, at least for now.

You are applying cosine to some "angle",
\(\large\rm cos(\color{orangered}{angle})=0\)

So which angle does this correspond to?
Think of your special angles.
Cosine of what angle is 0?

Answer 28

oo it also tells me to replace the "8x" with just a or something

Answer 29

cosine of 0 is 0!

Answer 30

Woops, cosine of 0 is 1 :O
That's not the angle we wanted.

Answer 31

I made that mistake on your last question lol

Answer 32

cos(0) = 1

Answer 33

Ya we don't care about that, we have cos(angle)=0.
We don't care about cos(angle)=1.

It's up at pi/2, ya?

Answer 34

yes,

Answer 35

So \(\large\rm cos(\color{orangered}{angle})=0\) tells us that

\[\large\rm \color{orangered}{angle}=\frac{\pi}{2}\]

Answer 36

Or more accurately, for our problem,\[\large\rm \color{orangered}{8x}=\frac{\pi}{2}\]

Answer 37

cos(x) is NOT same as cos*(x)

Answer 38

cos(x) = 0

can you tell me again how/why you're allowed to use zero product property here ?

Answer 39

Well.. I don't know, exactly. When should we use the zero product property?
Anytime we just move all the terms to one side, shouldn't the other side be 0, thus we will be using the property.

Answer 40

There's many trig equations.. This is just ONE of the ways to solve them, right? I need to memorize all the formulas? Man..

Answer 41

You don't have to memorize a lot of formulas. Let's first understand zero product property and see why we cannot use it to solve cos(x) = 0.

Answer 42

\[x = \frac{ \pi }{ 16} +2\pi(n)\]

Answer 43

0 is the value of sin.. I don't think we can use it here because it is not an... "equation"?? I don't know a better word

Answer 44

cos(x) is ONE SINGLE FUNCTION.


you cannot break it into two pieces : cos and (x)

Answer 45

\[\large\rm 8x=\frac{\pi}{2}+n \pi\]Woops, don't forget to divide the other part by 8 as well :)\[\large\rm x=\frac{\pi}{16}+\frac{n \pi}{8}\]

Answer 46

Yes, I have the answer, but I'm not fully understanding the steps, yet.

If you mind..
......Can we do this all over again

Answer 47

and this is just only one of the many ways to solve it :( there's so many diff equations to solve with diff methods

Answer 48

Ya :D Trig is awesome, amirite? :D

Remind me of a Mario game,
you're trapped in a room full of pipes...
and they all connect somehow :d

Answer 49

Gog gog gog gog gaggag gag

Answer 50

derivatives vs trig??

Answer 51

calculus + trig is where it's at :D
It's all down hill after that lol

Answer 52

Are you familiar with twiddla?
It'd be so much easier if I can just draw some of this out... instead of typing again t.t

Answer 53

I can go on the website, right?

Answer 54

https://www.twiddla.com/a0a4wk