Question

Find "x" in
3 sin(x)-sqrt(3) cos(x) = sqrt(3)
Please show working out. Thanks! Find "x" in
3 sin(x)-sqrt(3) cos(x) = sqrt(3)
Please show working out. Thanks! @Mathematics

Answer 1

you have to use a calculator, its a non linear equation ,

Answer 2

3sin(x) - sqrt(3)cosx - sqrt(3) = 0

Answer 3

But we're not allowed to use calculators to solve this.

Answer 4

3sin x = sqrt 3 + sqrt 3 cos x

Answer 5

3 sin x = sqrt 3 ( 1 + cos x )

Answer 6

sqrt 3 sin x= 1 + cos x

Answer 7

I think we have to solve it by R-method (Auxiliary Angle Method) but it doesn't work when I try it.

Answer 8

whats R method?

Answer 9

umm... do you know sum of products of trigonometric functions?

Answer 10

ummm.. yeah i invented them

Answer 11

lol well it uses that and then somehow you let 3 equal to cos(u) and sin(u) equal to sqrt3... wait look on this site. I can't explain it properly
http://trevorpythag.co.uk/2009/mathematics/trigonometry/auxiliary-angle-method-for-solving-trigonometry-equations/

Answer 12

ok im reading this , i invented this so its not going to be hard to remember

Answer 13

ok. =D

Answer 14

asinx - bcosx = rsin(x-a)

Then find r = sqrt(a^2+b^2)

Answer 15

yup I did that. It equals to
2sqrt(3)

Answer 16

then find tan= a/b

Answer 17

tan(a) = a/b*

Answer 18

1/sqrt(3) ?

Answer 19

3/sqrt(3) i think

Answer 20

ok
... but wouldn't we make tan(x) by doing b/a?

Answer 21

ok we will rewrite the left side of the equation
3 sin(x)-sqrt(3) cos(x) =r sin(x+y)
3 sin(x)-sqrt(3) cos(x) = r sinx cosy + r siny cos x
so
rcos y = 3, r sin y = -sqrt 3

Answer 22

woops tan(x) = b/a -> sqrt(3)/3 ->30

Answer 23

@perl why does rsiny=-sqrt3? Why is sqrt3 negative?

Answer 24

using the identity, cos^2 + sin^2 = 1, we have
[ r cos y ] ^2 + [ r sin y ] ^2 = r^2,
3^2 + (-sqrt 3)^2 = r^2

Answer 25

because we are matching the left and right terms

Answer 26

?

Answer 27

if it is true that
3 sin(x)-sqrt(3) cos(x) = r sinx cosy + r siny cos x
then it must be true that
3 sin x = r sin x cos y
-sqrt 3 cos x = rsin y cos x

Answer 28

There's a easy way to do this. You just substitute:

\[sinx = \frac{2t}{1+t^2} , cosx = \frac{1-t^2}{1+t^2}\]

Getting =

\[6t-\sqrt3 + \sqrt3t^2 = \sqrt3 +\sqrt3t^2 \rightarrow 6t = 2\sqrt3 \rightarrow t = \frac{\sqrt3}{3} \rightarrow \tan(x/2) = \frac{\sqrt3}{3} \rightarrow \]

x = \[1/3 (π+6 n π)\]

Answer 29

if it is true that
3 sin(x)-sqrt(3) cos(x) = r sinx cosy + r siny cos x
then it must be true that
3 sin x = r cos y * sin x
-sqrt 3 cos x = rsin y cos x

Answer 30

and π/3+2 n π

Answer 31

its like equating coefficients ,

Answer 32

alfie, where did you get that identity?

Answer 33

@alfie Thanks! but that's using T-method. I have to solve this question using R-method

Answer 34

or that substitution?

Answer 35

@perl ok I get why now!

Answer 36

@perl http://www.qc.edu.hk/math/Certificate%20Level/t%20method.htm

Answer 37

hmmm, thats a weird substitution. its true that (2t/(1+t^2)) + ( 1-t^2 /1+t^2)^2 = 1 , but ...

Answer 38

duplicity, now
rcosy = 3 , and r sin y = -sqrt 3 ,

Answer 39

@ alfie, what would x equal at the end if it was in degrees?

Answer 40

@perl ok continue on
=D

Answer 41

you do realise you could solve this by substitution, right?

Answer 42

and we know
using the identity, cos^2 + sin^2 = 1, we have
[ r cos y ] ^2 + [ r sin y ] ^2 = r^2,
3^2 + (-sqrt 3)^2 = r^2 ,
so r = sqrt 12

Answer 43

agree, what kind of substitution?

Answer 44

@agreene yes, but the question said to answer using R-method

Answer 45

@perl yup r=2sqrt3

Answer 46

r= sqrt(12)
a = 30

Therefore, sqrt(12(x+3)= sqrt(3)

sin(x+30) = sqrt(3)/sqrt(12)

r.angle = sin^-1=(sqrt(3)/sqrt(12)
= 30

x + 30 = 180 + 30 , 360-30

x + 30 = 210 , 330

x =...

Answer 47

now r cos y = 3, so 2sqrt 3 cos y = 3 , so cos y = 3/ ( 2 sqrt 3 ) = sqrt 3 / 2

Answer 48

so y = pi/6 + 2pi*n, 11pi/6 + 2pi*n

Answer 49

@ perl what's 'n'?

Answer 50

any integer, { ...-3,-2,-1,0, 1 , 2 ,...}

Answer 51

@mimi, Thanks! though... it's hard to read TT_TT

Answer 52

from:

3sin(x)-sqrt(3)cos(x)-sqrt(3)=0
Let
u=tan(x/2)
sin(x) = 2u/(u^2+1)
cos(x)= (1-u^2)/(u^2+1)

then we are at:
(6u)/(u^2+1)-(sqrt(3)(1-u^2))/(u^2-1)-sqrt(3)=0
do some rearranging:
\[-\frac{2(\sqrt{3}-3u)}{u^2-1}\]
solving that we get to
\[u=\frac{1}{\sqrt{3}}\]
sub that kid:

\[\tan\frac{x}{2}=\frac{1}{\sqrt{3}}\]
inverse tan both sides then multiply by 2
\[\large x=\frac{\pi}{6}\]

Answer 53

eep! so many answers... so x = 30, 300, 180?

Answer 54

lols yeah they left it in radian form. Not sure if my one is right though, it has been a while that i have done it.

Answer 55

my answer is in radians, so 30 degrees, btw.

Answer 56

but it isnt entirely complete, now that I think of it... there should be infinite solutions...
so yeah

Answer 57

is my one one right ? >_<

Answer 58

my one*

Answer 59

hmm.... o...k... I don't know!!

Answer 60

oh and @mimi isn't it sin(x-30) not sin(x+30)

Answer 61

solutions cannot be just one angle, they have to be infinite since the tangent is a periodic function, you will have to plug in the period, otherwise that's not correct, or partially correct ;)

Answer 62

something is wrong with my math, i got
now r cos y = 3, so 2sqrt 3 cos y = 3 , so cos y = 3/ ( 2 sqrt 3 ) = sqrt 3 / 2
y = pi/6 + 2pi*n, 11pi/6 + 2pi*n

Answer 63

\[x=2\pi n+\pi\]
And\[n \in\mathbb{Z}\]
should be your answer in radians (from my above solution)

Answer 64

but if you do
rcos y = 3, r sin y = -sqrt 3, then tan y = -sqrt 3 / 3 , that gives me

Answer 65

i dont get pi/6 as a solution

Answer 66

@Agreene I agree with you, I did your same reasoning and got the same solutions up there ;)

Answer 67

hmm... some how I got tan y = 1/sqrt3

Answer 68

if i do tan y = -sqrt 3 /3 , the solutions are 5pi/6 + pi*n,
but if i do
r cos y = 3, so 2sqrt 3 cos y = 3 , so cos y = 3/ ( 2 sqrt 3 ) = sqrt 3 / 2
y = pi/6 + 2pi*n, 11pi/6 + 2pi*n , i get a different answer

Answer 69

@alfie, I musta missed it in my awesome scrolling and wondering what everyone was doing O.o

Answer 70

eep! I'm confused

Answer 71

im getting different solutions for y

Answer 72

ok suppose you are given r cos y = 3, r sin y = -sqrt 3 , r = 2 sqrt 3 , you can divide
rsin y by r cos y , and get tan y = -sqrt 3/ 3 , or you can solve r cos y = 3 seperately

Answer 73

im going to have to post this as a new question, i found a hole in mathematics

Answer 74

@ perl Wholey!

Answer 75

one sec im going to research this

Answer 76

mimi, can you help ?

Answer 77

suppose I am given sin y = -1/2 and cos y = sqrt(3)/2, I get two different solutions.
if i do siny / cos y = tan y = (-1/2)/ (sqrt(3)/2)
tan y = -sqrt 3 / 3 , i get 5pi/6 + pi*n.
But if i do cos y = sqrt 3 / 2 , i get a different answer

Answer 78

lols, i have no idea when you're doing.
You're just making it more complicated.

Answer 79

i need to straighten out something first

Answer 80

its a side problem

Answer 81

I don't think that the question is that complicated :/

Answer 82

mim, its just a different question entirely.
suppose you want to solve the system of equations
sin y = -1/2
cos y = sqrt(3)/2 ,
it would be wrong to do tan y = -sqrt 3 / 3 , that comes out with wrong solutions

Answer 83

woudn't it be -2/2sqrt(3) ?

Answer 84

(-1/2) / ( sqrt(3)/2) = -sqrt (3)/3