Question

*Find all solutions in the interval 0 < theta, 2pi for tan^2theta-3=0, and 2 sintheta+ sqrt3=0 . I'm so confused, HELP!

Answer 1

\[\tan^2(x)=3\]
First take square root of both sides, you will end up with two equations to solve after that.

Answer 2

Then use the unit circle.

Answer 3

\[\tan \theta=\sqrt{3}\]?

Answer 4

that is one equation

Answer 5

the second equation is tan(x)=-sqrt(3)

Answer 6

\[\tan^2(x)=3 \implies \tan(x)=\sqrt{3} \text{ or } \tan(x)=-\sqrt{3}\]

Answer 7

hint:
recall:
\[\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\sqrt{3}\]

Answer 8

Oh okay! So the solutions would be \[\pi/3, 2\pi/3, 4\pi/3, and 5\pi/3?\]

Answer 9

great

Answer 10

second one isolate the sin(x) part

(also note: I'm using x instead of theta because it is just easier to type)

Answer 11

Yes I figured :) Would the 2nd equation would be \[\sin \theta= -\sqrt{3}/2?\]

Answer 12

yep!

Answer 13

Oh thank you sooo much!!!

Answer 14

you got it from there?

Answer 15

I believe so... then the sin values would be ... 5pi/ 3, and 4pi/3?

Answer 16

right x or you know theta would be 5pi/3 or 4pi/3
good job
you weren't that confused

Answer 17

I know the unit circle, but these equations just look confusing.

Answer 18

you remember solving linear equations in algebra right?

Answer 19

Yes

Answer 20

like you know how to solve something like
2x+5=4 for x?

Answer 21

it is the same way
you isolate the trig function
then you solve for the variable inside the trig function

Answer 22

I wish my teacher would just explain it like that. Would make it much clearer, lol

Answer 23

if this makes more sense to you replace tan(theta) with p
so the first equation you first solve for p: p^2-3=0
p^2-3=0
add 3 on both sides
p^2=3
take square root of both sides
p=sqrt(3) or p=-sqrt(3)
then replace p with tan(theta)
tan(theta)=sqrt(3) or tan(theta)=-sqrt(3)

second equation replace sin(theta) with p and solve for p:
2p+sqrt(3)=0
subtract sqrt(3) on both sides
2p=-sqrt(3)
divide both sides by 2
p=-sqrt(3)/2
and so on...

Answer 24

Yes, much easier! I thought maybe I had to use an identity or something

Answer 25

now I do admit this trig equations can get more complicated

Answer 26

The next one is a little different. \[\cos (3\theta)=+\sqrt{3}/2\]

Answer 27

that one is also not too bad...
is the interval the same for the solution?

Answer 28

yes 0 to 360

Answer 29

just divide both sides by 3?

Answer 30

\[\text{ so we want to solve for } \theta \text{ such that } 0 < \theta <360 \\ \text{ but first we will need to solve for } 3 \theta, \text{ let's call this } p \\ \text{ so we need to figure out the solution interval in terms of } p \text{ now } \\ p=3 \theta \implies \frac{p}{3}=\theta \\ \text{ so } 0 < \theta < 360 \implies 0 < \frac{p}{3} < 360 \implies 0(3) <p<360(3) \\ \implies 0<p<1080\]
So you first need to solve
\[\cos(p)=\frac{-\sqrt{3}}{2}\]
on the interval (0,1080)

Answer 31

that is only 3 rotations around the circle

Answer 32

find the solutions in the first rotation
then we will just add 360 to get the solutions in the second rotation
and then we will add 720 to get the solutions in the third rotation

Answer 33

once we find p
we will be able to find theta easily

Answer 34

uhm... pi/6, and 5pi/6?

Answer 35

in the first rotation solutions should be 5pi/6 , 7pi/6

Answer 36

oops but we should probably use degrees since our interval is in terms of degrees

Answer 37

oooh

Answer 38

so first rotation solutions should be 150 and 210

Answer 39

why not 35?

Answer 40

35? you mean 30? we are looking for when cosine value is negative sqrt(3)/2 not positive
second rotation solutions should be 150+360 and 210+360
third rotation solutions should be 150+720 and 210+720

Answer 41

Why negative cosine?

Answer 42

doesn't your equation say:
\[\cos(p)=-\frac{\sqrt{3}}{2} ?\]

\[p=150,210,510,570,870,930 \\ \text{ remember } p=3 \theta \\ 3 \theta=150,210,510,570,870,930\]

Answer 43

err I'm blind I thought I seen a negative in front

Answer 44

you did have a positive

Answer 45

hahah

Answer 46

but maybe I should finish solving this one to give you an idea for the positive version of sqrt(3)/2 for cos

Answer 47

perfect!

Answer 48

\[p=150,210,510,570,870,930 \\ \text{ remember } p=3 \theta \\ 3 \theta=150,210,510,570,870,930\]
last step is divide both sides by 3

Answer 49

by the way this is called the lazy way of writing 6 different equations :p

Answer 50

\[\theta=\frac{150}{3}, \frac{210}{3}, \frac{510}{3}, \frac{570}{3}, \frac{870}{3}, \frac{930}{3}\]

Answer 51

and you do the divisions and you are done

Answer 52

but anyways let's go back to the equation you actually had

Answer 53

awesome! wow.. wouldn't have thought to do that

Answer 54

\[\cos(3 \theta)=\frac{\sqrt{3}}{2}\]
\[\text{ where} 0 < \theta<360\]

Answer 55

same thing
I would replace 3theta with some other variable like p
and then write the solution interval in terms of p as well so I know which p to find

Answer 56

okay, so 30, and... 330?

Answer 57

right

Answer 58

we will have 2 more rotations of solutions to look at

Answer 59

30
330
30+360
330+360
30+2(360)
330+2(360)

Answer 60

now these will be the solutions for p

Answer 61

and you have p=3 theta
so just divide your solutions for p by 3 and you are done

Answer 62

Wow, thank you so much. I've never been shown that way before

Answer 63

well I hope it is easier for you than the way you have been shown
curious what way you have been shown though

Answer 64

For some, they used factoring. Which confused me more

Answer 65

I don't how you would do factoring for that one...
but I do see how one might want to factor:
\[\cos^2(x)-2\cos(x)-3=0 \\ \text{ factoring left hand side gives } \\ (\cos(x)-3)(\cos(x)+1)=0\]
set both factors equal to 0
you know since a*b=0 then this means a=0 or b=0

Answer 66

\[\cos(x)-3=0 \text{ or } \cos(x)+1=0 \\ \cos(x)=3 \text{ or } \cos(x)=-1 \\ \text{ no solutions for } \cos(x)=3 \\ \text{ but you will have solutions for } \cos(x)=-1 \]

Answer 67

Yeah it was for this one I had \[2\cos^2 \theta-1=0\]

Answer 68

actually let me try something different for the one you said they factored on...
\[\cos(3x)=\frac{\sqrt{3}}{2} \\ \cos(3x)=\cos(2x+x)=\cos(2x)\cos(x)-\sin(2x)\sin(x) \\ \cos(3x)=(\cos^2(x)-\sin^2(x))\cos(x)-(2 \sin(x) \cos(x)) \sin(x) \\ \cos(3x)= \cos^3(x)-\sin^2(x) \cos(x)-2 \sin^2(x) \cos(x) \\ \cos(3x)=\cos^3(x)-(1-\cos^2(x)) \cos(x)-2 (1-\cos^2(x)) \cos(x) \\ \cos(3x)=\cos^3(x)-\cos(x)+\cos^3(x)-2 \cos(x)+2 \cos^3(x) \\ \cos(3x)=4 \cos^3(x)-3 \cos(x) \\ \text{ so you have the equation } \\ 4 \cos^3(x)-3\cos(x)=\frac{\sqrt{3}}{2}\]
that is a monster

Answer 69

oh they factored on 2 cos^2(x)-1=0
you can factor on that one or just solve for cos^2(x) first then take square root of both sides
either way

Answer 70

if the factoring bit confused with the trig function
replace the trig function with some variable as mentioned before

Answer 71

Holy moly, yeah just looks like a lot more work. Oh true, thats a good idea

Answer 72

\[2p^2-1 =(\sqrt{2} p)^2-(1)^2=(\sqrt{2} p-1)(\sqrt{2}p+1)\]

Answer 73

\[2 \cos^2(x)-1=(\sqrt{2} \cos(x)-1)(\sqrt{2} \cos(x)+1)\]

Answer 74

anyways I have to leave for now

Answer 75

make a new post if you need anymore help

Answer 76

Thank you so much!!

Answer 77

np :)