Question

More Trigonometric Equations! :) .. :'(
Solve for (0,2pi):
sin2xsinx+cos2x+cosx=1

if you could explain the steps too that would be AMAZING! :)

Answer 1

please.. im struggling i wasn't in class for this lesson..

Answer 2

Do your know the double angle formula

Answer 3

yea i know all the identities

Answer 4

why dont u substitute for sin 2x and cos 2x

Answer 5

i know you chance the sin2x and cos2x with 2sinxcosx and cos^2x-sin^2x but what do you do then?

Answer 6

sin2xsinx+cos2x+cos=1
2sinxcoxsinx + 1-2sin^2 x+cos x =1
?

Answer 7

uhhh wait up ..one second lemme do that.

Answer 8

Oh boy this one is a doozy :u

Answer 9

tell me about it :/

Answer 10

wait how did you get that @thushananth01 ?

Answer 11

He went offline :c sec I can help.

Answer 12

okay thank you! :)

Answer 13

\[\Large\rm \color{royalblue}{\sin2x} \sin x + \color{orangered}{\cos2x} + \cos x =1\]So we need to apply our Double Angle Identities to these two colored parts, yes?

Answer 14

yea 2sinxcosx and cos^2 x-sin^2 x

Answer 15

Ok good.
Do you remember the other forms for the Cosine Double Angle?
Because we want to use this one:\[\Large\rm \color{royalblue}{\sin x \cos x} \sin x + \color{orangered}{1-2\cos^2x} + \cos x =1\]

Answer 16

(In orange)

Answer 17

oh wait

Answer 18

i wrote the question wrong there is not addition sign between cos2x and cosx

Answer 19

im sorry! :/

Answer 20

Multiply?

Answer 21

yeap

Answer 22

Mmm ok sec, let's start over.
\[\Large\rm \color{royalblue}{\sin2x} \sin x + \color{orangered}{\cos2x}\cos x =1\]

Answer 23

I posted my Sine Double Angle wrong >.< my bad.

Answer 24

okay

Answer 25

its okay! i saw lol :)

Answer 26

\[\Large\rm \color{royalblue}{2\sin x \cos x} \sin x + \color{orangered}{(\sin^2x-\cos^2x)}\cos x =1\]

Answer 27

correct

Answer 28

do i cancel out the sinx with the 2sinx?

Answer 29

No, you multiply them together, giving you a sin^2x, yes?\[\Large\rm 2\sin^2 x \cos x + \color{orangered}{(\sin^2x-\cos^2x)}\cos x =1\]

Answer 30

yeap that looks right

Answer 31

Umm this is a little tricky.
Let's seeeeee....

I guess we'll write everything in terms of cosine.
So rewrite our sin^2x as (1-cos^2x) in both places.

Answer 32

\[\Large\rm 2(1-\cos^2x) \cos x + \color{orangered}{(1-\cos^2x-\cos^2x)}\cos x =1\]

Answer 33

I know I know I'm stealing all the fun >.< lol sorry
I like my equation tool hehe

Answer 34

lol its all good :)

Answer 35

So then we combine like terms, expand everything out, then combine like terms again -_-

Answer 36

\[\Large\rm 2(1-\cos^2x) \cos x + \color{orangered}{(1-2\cos^2x)}\cos x =1\]\[\Large\rm 2\cos x-2\cos^3x+\cos x-2\cos^3x =1\]\[\Large\rm 2\cos x-2\cos^3x+\cos x-2\cos^3x -1=0\]Oh boy this isn't working out nicely.. hmmm thinking +_+

Answer 37

jeez i didnt think it would take this long for this problem. maybe is was ment to be written in terms of sine?

Answer 38

Probably not, since there is a cosx multiplying each term.
There wouldn't be a nice way to deal with those.

Answer 39

oh do you think you can get the answer? :o

Answer 40

Yah I'll throw it into Wolfram, that should help me see what's going on here >.<

Answer 41

okay :) cool

Answer 42

Ohhh I'm so dumb.
Yah I forgot to "rethink the problem" when we made the adjustment.

This is just simply the Sine Additive Identity.\[\Large\rm \sin(a+b)=\sin a \cos b + \sin b \cos a\]That look familiar?

Answer 43

ohhhhhh snap ! i feel even stupider lol

Answer 44

Do you see it popping out at you? :O

Answer 45

im going to erase my work and start over :) can i tell you if i need help? :/

Answer 46

Sure :U
I'm hittin the sack soon, but I'll be on for another 20 prolly.

Answer 47

okay ill try to work fast

Answer 48

Woops woops woops.

We have the form sin2xsinx + cos2xcosx

I should have posted the Cosine Additive Identity, not the Sine.

\[\Large\rm \cos(a-b)=\cos a \cos b+\sin a \sin b\]

Answer 49

are you sure its not the cosine sum identity?

Answer 50

>.<

Answer 51

ohhh okay lolol

Answer 52

And it looks like it's the difference, not the addition one, right?
Did I get that correct?
Yah I think my sign is ok.

Answer 53

yea your right

Answer 54

hey im stuck >.< @zepdrix

Answer 55

i dont see it poping out .. -_-

Answer 56

So our formula is:
\[\Large\rm \cos\left(\color{royalblue}{a}-\color{orangered}{b}\right)=\cos\color{royalblue}{a}\cos\color{orangered}{b}+\sin\color{royalblue}{a}\sin\color{orangered}{b}\]And our problem is:\[\Large\rm \cos\color{royalblue}{2x}\cos\color{orangered}{x}+\sin\color{royalblue}{2x}\sin\color{orangered}{x}=1\]yes?
Is it popping yet? :U

Answer 57

isnt x just equal 1?

Answer 58

Whut? +_+

Answer 59

im so lost...

Answer 60

im so sorry..

Answer 61

im realllly dumb.. scratch that.

Answer 62

Try to read what I did with the colors.
Match up the colors.

Answer 63

i put cos(2x-x)=cos2xcosx+sin2xsinx

Answer 64

Ok great!
So the left side of the equation simplified!\[\Large\rm \cos\left(\color{royalblue}{2x}-\color{orangered}{x}\right)=1\]

Answer 65

So we get cos(x)=1, yes?
Not x=1 silly billy +_+

Answer 66

yeaaa lol my baddd so that means its 0 degrees and 180 degrees and 360 degrees right? then make it into radians?

Answer 67

No, just 0 and 360.
Not 180.
Cosine at 180 is -1 yes?

Answer 68

ohhh yeaa right

Answer 69

But careful!
Look back at your interval, did you post it correctly?
Or should one of the sides have square bracket?

Answer 70

Because \(\Large (0,2\pi)\) gives us no solutions.

Answer 71

well its 0 degrees greater or equal to x greater or equal to 360 degrees

Answer 72

no wait less then or equal to

Answer 73

\[\Large\rm 0\le x\le 360\]Like that?

Answer 74

yaaa sorr i could only do > and < lol

Answer 75

i cant make that line under it lol

Answer 76

In equation tool it's \le
short for less than :P
Kinda hard to remember all those silly shortcuts though hehe

Answer 77

Ok if we have the "or equal to" on both equalities,
then both sides are square brackets.

\(\Large\rm [0,360]\)

Answer 78

So we include both end points.
So yes, 0 and 360.

Answer 79

so i leave it in degrees right?

Answer 80

It doesn't matter.
Since the problem was given in degrees, yes it's a good idea to leave it in degrees.

Answer 81

okay THANK YOU JESUS zepdrix thank you thank you thank you i love you

Answer 82

Lol -_- np.

Answer 83

man i have a pre calc text tomorrow.. wish me luck

Answer 84

Ooo fun \c:/ Good luck!

Answer 85

Trig is the best most fun subject ever :U
Don't slack off.

Answer 86

It's full of amazing identities and great stuff that you'll never ever use unless you decide to become a math major! :D

Answer 87

thanks. :) like i understand it i just get really unsure about my problem solving so then i get unsure so then i always feel like im doing it wrong

Answer 88

Mm I'm sure you'll do great c:

Answer 89

i have a B my grade depends on this test and a final on tuesday.. *gulp*