Question

Solve 4cos^2A = 3cosA for 90 13 years ago

Answer 1

put it all on one side, and factor:
\[4\cos^2 A- 3 \cos A = 0\]\[\cos A (4 \cos A - 3) = 0\]
so cos A = 0, and cos A = 3/4
cos A = 3/4 cannot exist, since cos A is negative in the second quadrant.
so cos A = 0 at 90º

Answer 2

Oh okay, I was getting confused becayse of the 4 and 3 infront of cos Lol

Answer 3

Wait, I'm a little confused. cos A = 0 and cos A = 3/4

Answer 4

yes, it is like a quadratic equation with more than one value of x. only it cannot be 3/4.

Answer 5

Oh, right. Lol Sorry

Answer 6

I still don't really understand what to do. I know it's plain and simple :p But I'm just not following

Answer 7

I get how to move everything to the other side, but from there on out, I'm a little confused

Answer 8

It might be easier to let z = cos(A). So z^2 = cos^2(A)


This means

4cos^2A = 3cosA

is the same as

4z^2 = 3z

Answer 9

Solve that equation for z. Once you have solutions for z, use them to find solutions for A.

Answer 10

Oh, like shortening it?

Answer 11

yes

Answer 12

DOes this look right so far?
4z^2 - 3z = 0
(z + 3)(z - 1)

Answer 13

Or wait... no Lol that doesn't look right

Answer 14

all you need to do is factor a 'z' from 4z^2-3z to get z(4z - 3)


So

4z^2-3z = 0

becomes

z(4z - 3) = 0

Answer 15

Then break it up to get z = 0 or 4z - 3 = 0

Answer 16

How would you break it up?

Answer 17

Do you mean like simplify so it's just z = 0?

Answer 18

well if AB = 0, then either A = 0 or B = 0 right?

Answer 19

True, yes

Answer 20

So we're applying that to break up

z(4z - 3) = 0

into

z = 0 or 4z - 3 = 0

Answer 21

Remember we let z = cos(A), so if z = 0, then cos(A) = 0

Answer 22

Also, if 4z-3 = 0, then z = 3/4 and cos(A) = 3/4

Answer 23

(sorry if I'm asking so many questions) How did z = 3/4? I see that 4z - 3 = 0, does that relate to how z = 3/4?

Answer 24

sry I jumped a few steps

4z - 3 = 0

4z - 3+3 = 0+3

4z = 3

4z/4 = 3/4

z = 3/4

Answer 25

Oooh, alrighty (: I see now

Answer 26

This might clear some air for me; what is it exactly that the question is asking me to look for?

Answer 27

You're looking for the value of A that makes 4cos^2(A) = 3*cos(A) true

Keep in mind that 90 <= A <= 180

Answer 28

Okay, hold on

Answer 29

ok

Answer 30

I'm not sure where else to take it from here. I know cosA = 3/4, but that's not what A is right?

Answer 31

no, that's what the cosine of A is

Answer 32

Right

Answer 33

if cos(A) = 3/4, then you can take the arccosine of both sides to eliminate that "cos" on the left side

Answer 34

so A = arccos(3/4)

Answer 35

So like cosine inverse? Where's it's like \[\cos^{-1} \]

Answer 36

yes it's the inverse of cosine

Answer 37

\[\Large \arccos(x) = \cos^{-1}(x)\]

Answer 38

The left side "arccos" is used more because cos^(-1) may be confused with 1/cos

Answer 39

Alright, so I put \[\cos^{-1} (3/4)\] and got 41.4

Answer 40

good

Answer 41

is that between 90 and 180?

Answer 42

Oh right, because 1/cos is sec

Answer 43

yes

Answer 44

No that is not

Answer 45

so that's not the solution

Answer 46

since it's not in the desired interval

Answer 47

now solve cos(A) = 0

Answer 48

Alrighty, I got 90 as my answer

Answer 49

good, so A = 90

Answer 50

which is in the interval 90 <= A <= 180

Answer 51

Ahh, I see (: Thank you!!!

Answer 52

yw

Answer 53

If the interval was 90 < A < 180, then it'd be a different story. However, we're including the endpoints.

Answer 54

So when it's <= it includes endpoints?

Answer 55

yes "<=" means "less than or equal to"

Answer 56

Yeah (: Why is it that when it's just less than or more than, it doesn't include the endpoints?

Answer 57

Because when you say "less than 5" for instance, you mean "any number that is below 5 but NOT including that number 5"

Answer 58

Oh okay (: Makes sense. Thank you again, I feel bad, you've been having to help me a lot

Answer 59

but when you say "less than OR equal to 5" you mean " something smaller than 5 OR something equal to 5"

Answer 60

no it's perfectly fine as long as you're getting it?

Answer 61

Oh okay, it stops at a certain point. It's not infinite. And yes, I understand better now (:

Answer 62

that's great :)