Question

How would I isolate the trig function here..?

Answer 1

\[\sqrt{3}\cos x \tan x-\cos x=0\]

Answer 2

@Nnesha

Answer 3

@jim_thompson5910

Answer 4

you can factor out a cos(x) and then use the zero product property



\[\Large \sqrt{3}\cos(x)\tan(x) - \cos(x) = 0\]

\[\Large \cos(x)(\sqrt{3}\tan(x) - 1) = 0\]

\[\Large \cos(x)=0 \text{ or } \sqrt{3}\tan(x) - 1 = 0\]

\[\Large \cos(x)=0 \text{ or } \sqrt{3}\tan(x) = 1\]

\[\Large \cos(x)=0 \text{ or } \tan(x) = \frac{1}{\sqrt{3}}\]

\[\Large \cos(x)=0 \text{ or } \tan(x) = \frac{\sqrt{3}}{3}\]

I'll let you finish

Answer 5

Oh...you solve for both tan and cos here?

Answer 6

you solve for x ultimately

Answer 7

tan and cos are functions

you want the value(s) of x that makes the original equation true

Answer 8

Hmm, so if cos=0,

general solution would be

pi/2+2npi?

Answer 9

and 3pi/2+2npi?

Answer 10

cos(x) = 0

cos = 0 makes no sense

Answer 11

cos(x)=0 when x=pi/2

Answer 12

or 3pi/2

Answer 13

yes, if cos(x) = 0, then x = pi/2 + 2pi*n or x = 3pi/2+2pi*n

Answer 14

those are not general solutions though right, just solving for x?

Answer 15

as least for cos

Answer 16

x = pi/2 + 2pi*n or x = 3pi/2+2pi*n

are the general solutions for cos(x) = 0

n is any integer

Answer 17

do the same for tan(x) = sqrt(3)/3

Answer 18

alright, that would be tan 30 degrees..

pi/6+npi

5pi/6+npi

Answer 19

good

Answer 20

uh...so which ones are correct?

Answer 21

they all are because they satisfy the original equation

Answer 22

they all collectively form the general solution

Answer 23

actually for tan pi/2, isn't that undefined?

Answer 24

@jim_thompson5910

Answer 25

good point, tan(x) = sin(x)/cos(x)

so if cos(x) = 0, then tan(x) is undefined

Answer 26

so the restriction is that \[\Large \cos(x) \neq 0\]

Answer 27

meaning that x can't be pi/2, can't be 3pi/2, etc

Answer 28

so just pi/6+npi and 5pi/6+npi?

Answer 29

so this is why it's good to check the original equation

Answer 30

yes

Answer 31

got it....thanks!

Answer 32

np