Question

Trigonometric Equation General Solution!!

Answer 1

\[If (1/6)sinx,cosx,tanx \] are in Geometric Progression

Answer 2

then find the general solution for x.

Answer 3

@Luigi0210 @mihirb @NoelGreco @@amistre64 @souvik @surjithayer @Fifciol @dumbcow @tiannaericka @jhonyy9 @AkashdeepDeb @amistre64

Answer 4

@ganeshie8

Answer 5

@Yttrium

Answer 6

Help Guys/Gals!!

Answer 7

\(\large \frac{\cos x}{\frac{1}{6}\sin x} = \frac{\tan x}{\cos x}\)


cross multiply and simplify a bit

Answer 8

get everything into cos

Answer 9

but how do I simplify cos^3x???

Answer 10

\[\cos^3x=\sin^2x/6\]

Answer 11

@ganeshie8

Answer 12

I will get a cubic equation,oi???

Answer 13

yes, dont get scared of cubics
change sin to cos

Answer 14

no,not scared.....Just how to proceed from

Answer 15

\[6\cos^3x+\cos^2x-1\]

Answer 16

shud I take cos common??

Answer 17

\[cosx(6\cos^2x+cosx-1/cosx)\]

Answer 18

nope. we going to solve the cubic in cosx straight.

Answer 19

how??

Answer 20

u familiar wid below :-
1) rational root theorem
2) synthetic division / long division

Answer 21

if so, we can solve it very easily without any pain

Answer 22

kinda ok with synthetic div....whats the first one??

Answer 23

can u help me with the synthetic division??

Answer 24

I am not getin any value satisfying this.......not 1,-1,2,-2

Answer 25

sure :)

rational root theorem says this :-
```
for any polynomial, rational roots will be of form p/q

where,
p = factors of constant term
q = factors of leading coefficient
```

Answer 26

\(6\cos^3x+\cos^2x-1 \)

say cosx = t

\(6t^3 + t^2 - 1\)

constant term = -1
leading coefficient = 6

Answer 27

so substitute -1/6???

Answer 28

no

Answer 29

and check if it satisfies??

Answer 30

not just -1/6

Answer 31

okay...then??

Answer 32

find all factors of constant term : -1
find all factors of leading coefficient : 6

take all combinations

Answer 33

O_o

Answer 34

read my reply on rational root theorem definition again ok

Answer 35

so 8 combinations??

Answer 36

factors of -1 : +1, -1
factors of 6 : 1, 2, 3, 6, -1, -2, -3, -6

Answer 37

oh I missed some

Answer 38

then take them in pairs??

Answer 39

after that??

Answer 40

so, possible rational roots : \(\pm 1, \pm \frac{1}{2}, \pm \frac{1}{3}, \pm \frac{1}{6}\)

Answer 41

why 1/2??1/3???

Answer 42

yes test them and see if u get lucky wid any one

Answer 43

cant it be 2,3??

Answer 44

or does it go both ways??

Answer 45

NOPE. read my reply on rational root theorem definition.

factors of -1 go in the top.
factors of 6 go in the bottom only

Answer 46

oh got it

Answer 47

cool :)

Answer 48

so I will substitute the values.....see which satisfies....and get back to u.that okay???

Answer 49

Yes. .. test all the listed possible values one by one until u find a root.
there is no easy way out; factoring is not that straightforward here.

Answer 50

I think its 2

Answer 51

nope........I know its 2

Answer 52

2 is not even in the list :|

Answer 53

yes it is.........Oops sry.I meant 1/2

Answer 54

good :) let me verify quick

Answer 55

Yes ! t = 1/2 is a root ! good job !!
wat did u get as the quotient ?

Answer 56

ta

Answer 57

what quotient???O-_o

Answer 58

quotient after synthetic division by 1/2

Answer 59

just hold on

Answer 60

okie take ur time

Answer 61

\[6x^2+4x+2\]

Answer 62

perfect !
check if that quadratic CAN have any real roots,
if it doesnt have real roots, its useless of rus. we can discard/ignore it.

Answer 63

u knw how to check if a quadratic have real/complex roots right ?>

Answer 64

nope

Answer 65

yes

Answer 66

we check the Determinant b^2-4ac

Answer 67

answer is nope

Answer 68

ohk i get u :)

Answer 69

so our only solution is t = 1/2

now substitute back the t = cosx

Answer 70

\[\sqrt{-32}\]

Answer 71

Okay...Now I get u

Answer 72

yes, we are not doing complex solution now. so we can discard it

Answer 73

oly real solution is t = 1/2

Answer 74

so ans is

Answer 75

u want ansurs ha

substitute t = cosx

cosx = 1/2

Answer 76

\[2n(\pi)\pm(\pi)/3\]

Answer 77

no I am sayin them

Answer 78

jk :)

\(x = \cos^{-1}(1/2)\)

Answer 79

\[2n(\pi)\pm5(\pi)/3\]

Answer 80

Excellent !!
two general solutions ok

Answer 81

are these the solutions??

Answer 82

sry....just replace pi/3 by pi/6

Answer 83

Yup !

Answer 84

I'll correct it here...

Answer 85

Thanks a TON

Answer 86

u were great.......

Answer 87

nope u were great !

wait a sec pi/3 = 60
so pi/3 and 5pi/3 are ur particular solutions ok

Answer 88

yaya........I sawthat...lol me

Answer 89

;)

Answer 90

hmmm.....

Answer 91

k....onto next....Ta....stay tuned...if I have some more..

Answer 92

sure :)