Question

can someone with trig experience check my work for sine2x cosine x + cosine2x sine x = .5sqrt(3)

Answer 1

posting now

Answer 2

hint
sin(a+b)=sin(a)cos(b)+cos(a)sin(b)

Answer 3

oh ok

Answer 4

one more page

Answer 5

\[\sin(2x+x)=\frac{\sqrt{3}}{2}\]
\[\sin(3x)=\frac{\sqrt{3}}{2}\]
so this is the equation you solved?

Answer 6

oh mine
it looks like you took a more complicated route

Answer 7

is it incorrect? what is the easier process? lol

Answer 8

what I did above
use the sum identity for sin

Answer 9

but I will look at yours

Answer 10

if you have a*b=0
then you can do the whole a=0 or b=0

Answer 11

you can't do if a*b=sqrt(3)/2
then one of the factors has to be sqrt(3)/2
because that isn't necessarily true

Answer 12

do you know how to solve sin(3x)=.5*sqrt(3) ?

Answer 13

you would first find what u satisfies the equation sin(u)=.5*sqrt(3)
by using unit circle

Answer 14

then
you would divide answer by 3
to get what x is
since u=3x

Answer 15

ok, let me try that

Answer 16

pi/6

Answer 17

do you want to post my work?

Answer 18

me*

Answer 19

you are looking for when the y-coordinate is sqrt(3)/2

Answer 20

for what angles is the y-coordinate sqrt(3)/2

Answer 21

so x is not equal to pi/6?

Answer 22

let me post my work so u can see what i did

Answer 23

alright

Answer 24

posting now

Answer 25

one min this site is not very well developed

Answer 26

yeah it has been pretty slow

Answer 27

i have to upload it to my gdrive and then post it

Answer 28

but do you get pi/6 for the answer

Answer 29

the y-coordinate is sqrt(3)/2 when you have the angles pi/3 or 2pi/3
and you can do +2npi if you wanted

Answer 30

\[u=\frac{\pi}{3}+2n \pi \text{ or } u=\frac{2\pi}{3}+2n \pi\]
but u=3x

Answer 31

so divide both sides by 3

Answer 32

3*3=9 not 6

Answer 33

lol wow ok

Answer 34

\[\frac{\pi}{3} \cdot \frac{1}{3}=\frac{\pi \cdot 1}{3 \cdot 3}=\frac{\pi}{9}\]

Answer 35

but anyways that is just one value that satisfies the equation

Answer 36

there are infinitely more solutions to this equation

Answer 37

so other that that, its good? pi/9 is one solution an then i do the same thing for sin 2x+x?

Answer 38

sin(3x)=sqrt(3)/2
gives us two solutions for 3x in the first rotation of a circle
3x=pi/3 or 3x=2pi/3 correct?

Answer 39

so the answer would be infinity many solutions, he asks for all real solutions.

Answer 40

first of all do you understand my solutions for 3x the first rotation ?

Answer 41

yes

Answer 42

and then for any more rotations
you can do +2 pi*n
2*pi carries us once around the circle
2*pi*2 carries us twice around the circle
2*pi*3 carries us three times around the circle
...
2*pi*n carries us n times around the circle where n is an integer

also sin and cos have period 2pi

----
so anyways we have
3x=pi/3+2pi*n
or
3x=2pi/3+2pi*n

Answer 43

and then to solve for x you divide both sides by 3

Answer 44

\[x=\frac{\pi}{9}+\frac{2 \pi n }{3} \\ x=\frac{2 \pi}{9}+\frac{2 \pi n}{3} \\ \text{ where n is an integer }\]

Answer 45

that gives you all the solutions

Answer 46

for sin(2x)cos(x)+sin(x)cos(2x)=sqrt(3)/2

Answer 47

@briensmarandache do you have any questions
did i like totally make you speechless

Answer 48

or typeless in this case

Answer 49

i feel like this is a stupid question so i was sitting on it for a bit

Answer 50

I'm sure whatever question you have isn't stupid

Answer 51

so if i did the same thing for sine 2x+x = sqrt(3)/2, i would get the same solution for x or a solution that looks different but means the same thing or something different all together?

Answer 52

for sin(2x+x)=sqrt(3)/2?

Answer 53

2x+x is 3x

Answer 54

sin(2x+x)=sqrt(3)/2 is the same equation as sin(3x)=sqrt(3)/2

Answer 55

and both of those are the same equation as sin(2x)cos(x)+cos(2x)sin(x)=sqrt(3)/2

Answer 56

i see now, ok. lol, it was a stupid question! thanks!!!

Answer 57

sin(a+b)=sin(a)cos(b)+cos(a)sin(b)
so
sin(a)cos(b)+cos(a)sin(b)=sin(a+b)
sin(2x)cos(x)+cos(2x)sin(x)=sin(2x+x)=sin(3x)
I used that a was 2x and b was x

Answer 58

So first step to solve this equation you have here is to realize you can use the sum identity for sin

Answer 59

what if I had this:
\[\cos(2x)\cos(x)-\sin(2x)\sin(x)=\frac{\sqrt{3}}{2}\]

Answer 60

that would be cos(x+y)

Answer 61

right and you would have this as a result:
\[\cos(2x+x)=\frac{\sqrt{3}}{2} \\ \cos(3x)=\frac{\sqrt{3}}{2} \\ 3x=\frac{\pi}{6}+2n \pi \\ 3x=\frac{-\pi}{6}+2n \pi\]
where n is integer

Answer 62

or instead of -pi/6 you could have said 11pi/6
which ever one
i like to use cos is even when I can
tht is
cos(x)=cos(-x)

Answer 63

anyways I'm glad you understand more

Answer 64

thanks this really helped a lot!!!!

Answer 65

have fun and trig should be a blast !

Answer 66

i will, i like it already. once i have a better understanding of everything im sure ill love it more lol