Question

Trigonometric Equation

tanA cosA + √(3) cosA = 0

Answer 1

One more @hartnn

Answer 2

tan(A) is sin(A)/cos(A).

Answer 3

Sorry.

Answer 4

No prob

Answer 5

So it shall be sin A + cosA= 0
@hartnn ?

Answer 6

and where did sqrt 3 go ?

Answer 7

Oh insert that before cos A
Sorry

Answer 8

What's next?

Answer 9

ok, divide both sides by 2

Answer 10

1/2sinA + sqr(3) / 2

Answer 11

Then?

Answer 12

(1/2)sinA + (sqr(3) / 2) cos A = 0

Answer 13

you know what is sin (x+y) =... ?

Answer 14

Nope

Answer 15

What's it?

Answer 16

Bring what hartnn wrote in terms of sinA/cosA.

Answer 17

You mean tan?

Answer 18

I really don know

Answer 19

*dont

Answer 20

Exactly. Bring sinA/cosA on one side.

Answer 21

How?

Answer 22

I dont get you?

Answer 23

1/2 sinA = -sqrt3/2 cosA

1/2 sinA/cosA = -sqrt3/2

sinA/cosA = -sqrt(3)

tanA = -sqrt3

Answer 24

tanA is -sqrt(3) for what A?

Answer 25

Is that even possible? Moving to the other side?

Answer 26

Yes. Why not? :D

Answer 27

Okay. It's the 1st time i saw that. Haha

Answer 28

What's next then?

Answer 29

As I asked, for what value of A is tan(A) equal to -sqrt(3)?

Answer 30

A=60,300?

Answer 31

Oh that's -
How will that be?

Answer 32

Hmm.

Answer 33

No solution?

Answer 34

There is, actually. tan is negative when sin and cos have opposite signs.
@hartnn Aliter solution?

Answer 35

you just can't cancel cos A

cos A (tan A + sqrt 3) = 0

cos A = 0
tan A = -sqrt 3

we need to solve these both

Answer 36

90 deg is not a solution. tanA will be undefined. :-/
Can cancel.

Answer 37

How's that?

Answer 38

ofcourse,
but thats true for this particular problem only.
in general, we can't do this.
we need to mention one solution as 90 and then discard it saying tan 90 is not defined.
we need to show that analysis

Answer 39

But yeah, the other solution is the same.

Answer 40

What's gonna happen now?

Answer 41

True, but we can write alongside that since cos cannot be zero, we can cancel.

Answer 42

yes, thats my point, that we need to mention.

Answer 43

Please elaborate later. Thanks. I really am lost and confused

Answer 44

Kaylala, tanA + sqrt3 = 0.

Answer 45

X=-sqrt3
Then what?
@ParthKohli

Answer 46

Please help @hartnn @ParthKohli

Answer 47

ok, so u have problem finding angles for which tan A = -sqrt 3 ?

Answer 48

Yea i do

Answer 49

How to do this?

Answer 50

@hartnn

Answer 51

the ratio of sin A to cos A should give you -sqrt 3

Answer 52

I havnt encountered such
What will it be. @hartnn

Answer 53

look for angles with

sin A = sqrt 3/2
cos A = -1/2

or

sin A = -sqrt 3/2
cos A = 1/2

Answer 54

I see. Then?
I dont know

Answer 55

So i am to pict the bottom sin and cos?

Answer 56

you will be able to find both the angles

Answer 57

A={60,300, 30,330}
?
@hartnn

Answer 58

Am i right.
?
@hartnn

Answer 59

Brb

Answer 60

This is what I got. Try to follow the steps.

Answer 61

its only 120 and 300

the other method i was proposing,

(1/2)sinA + (sqr(3) / 2) cos A = 0

let B = 60
sin 60 = sqrt 3/2
cos 60 = 1/2

cos 60 sin A + sin 60 cos A = 0

sin (A+60) = 0
A +60 = 0
A+60 = 180
A+60 = 360

A = -60 = 300
A = 180-60 = 120

Answer 62

Okay i'm really confused which among the 3 answers and solutions is correct.
@ParthKohli @hartnn @kurtsloane

Could anyone help please. Thanks :)

Answer 63

@Callisto
Help. I'm confused with all these

Answer 64

@phi @UnkleRhaukus @petiteme @Preetha

Answer 65

i see there are many methods

Answer 66

my favourite one is multiply by a half then use sum/difference trig formulas in reverse

Answer 67

But whose answer is correct? @UnkleRhaukus
In your opinion
Who got it right

Answer 68

Is:
A= 60,300,30,330
Or
A=60,300

Answer 69

well A=30° doesn't work,that would make LHS =2

Answer 70

and A=330° would make LHS=1

Answer 71

Oh okay. Thanks

Answer 72

tan(300°)=-√3

Answer 73

and tan 120 = -√3
so, A = {120,300}

Answer 74

You know that tan is negative when either sin or cos are negative, but not both (because tan = sin/cos). So, since tan of your angle is -sqrt(3), you know that the angle is in either quadrant II or quadrant IV, because quadrant II x is negative and y is positive, and quadrant 4, x is positive and y is negative. You can use rules like that to help you figure out where the angle should be and quick-check your answers. But yeah A = {120,300}