Question

Trigonometry help please :)

Answer 1

???

Answer 2

Solve the equation on the interval \[0 \le \theta < 2\pi\]
\[6\cos(2\theta)=18\cos \theta - 12\]

Answer 3

Sorry about that, I was trying to make it as neat as possible.

Answer 4

How do I figure something like this out. I know I need to use a double-angle formula to solve it.

Answer 5

what is cos (2theta) = ???

Answer 6

In other words, what's the formula for cos 2x?

Answer 7

there is a trig identity relating cos(2 theta) and a function of cos theta

Answer 8

1 - 2 sin^2 x ?

Answer 9

the question is interms of cos so you should use 2cos^2x -1
\[\large\rm \cos( 2 \theta) = 2\cos^2 \theta -1 \]

Answer 10

Ohhh ok got it! But how would I solve the original problem with that. I'm really confused today :/

Answer 11

6cos(2theta)=6(3cos theta -2) divide both sides by 6

cos(2theta)=3costheta -2

Answer 12

\[\large\rm 6 ~\color{red}{cos(2\theta)}=18\cos \theta - 12\]
\[\large\rm 6 ~\color{red}{(2cos^2 \theta -1)}=18\cos \theta - 12\]
now you can distribute parentheses by 6

Answer 13

Its all so clear! So then it would be
\[12 \cos^2 \theta-6 = 18 \cos \theta -12\]

Answer 14

it is same as simple algebraic equation
let cos theta = x
so \[\large\rm 6 ~\color{red}{(2 cos^2 \theta -1)}=18\cos \theta - 12\]
\[\large\rm 6 ~\color{red}{(2x^2-1)}=18x - 12\]
solve for x

Answer 15

Alright so I need to solve for theta. Looking at it as x is easier I'll admit.
\[12x^2 -6= 18x - 12\]\[12x^2 - 18x -18=0\]
Then solve it as a quadrtaic equation right?

Answer 16

quadratic*

Answer 17

there is a mistake how did you get -18 ?
\[12x^2 -6= 18x - 12\]\[12x^2 - 18x \color{Red}{-18}=0\]

Answer 18

in order to cancel out `-12` from right side we should add 12 both sides

Answer 19

Right! that should be a 12. My bad!

Answer 20

what would be the quadratic equation ??

Answer 21

Using the quadratic formula or factoring?

Answer 22

or either? /.\

Answer 23

\[12x^2 -6= 18x - 12\]
set it equal to zero
what would you get when u add 12 both sides ?

Answer 24

\[12x^2-18x +6=0\]

Answer 25

looks good :) can you solve for x ?? o^_^o

Answer 26

its up to you. if quadratic formula is ur fvt then use that

Answer 27

Ok so 12 * 6 is 72... factors of 72 that up to -18 would be (-6) and (-12)

Answer 28

I'm having an off day.. oh gosh how do I factor after that?

Answer 29

1ST) is there any GCF(greatest common factor )?
always take out the gcf first and then it would be easy to factor it

Answer 30

True! 6 is the GCF

Answer 31

yep o^_^o

Answer 32

so I have 2x^2 - 3x + 1 = 0

Answer 33

2(x-1)(x+2)

Answer 34

x = 1 and -2 ?

Answer 35

why did you put 2 at front ?

Answer 36

nevermind!

Answer 37

(2x-1)?

Answer 38

6(2x^2 - 3x + 1) = 0
thats what u get \[\rm 6\cancel{=}0 ~~~~~~~~~~~2x^2-3x+1=0\]

now factor the quadratic equation 2x^2-3x+1=0

Answer 39

so I'm left with (2x-1)(x+2)

Answer 40

x = 1/2 and -2

Answer 41

But if it needs to fall in the interval then in reality theta is equal to 1/2

Answer 42

right?

Answer 43

\(\color{blue}{\text{Originally Posted by}}\) @jazy
so I'm left with (2x-1)(x+2)
\(\color{blue}{\text{End of Quote}}\)
plx show ur work
:)

Answer 44

2x - 1= 0
2x = 1
x = 1/2

x + 2 = 0
x = -2

Answer 45

You deserve a medal for asking a good question. Lol

Answer 46

i meant how did you factor it ??

Answer 47

(2x-1)(x+2) FOIL to doulbe check ur answer
did you get 2x^2-3x+1 ?

Answer 48

Yes I did.. I've been doing the work on paper though... that's why I have put down every detail.

Answer 49

i'm sure you did.
but (2x-1)(x+2) isn't equal to 2x^2-3x+1
i was just trying to figure out the mistake..

Answer 50

I'm sorry I got very confused. It was really (2x - 1)(x -2)
Sorry I didn't come back /.\

Answer 51

:) ts okay and yes that's right :)

set it equal to zero
2x-1=0 and x-2 = 0
solve for x