Question

Solve for \(x\in R\) :

\[ e^{\tan x} + \sec x = \log ( \cot x )+ \csc x \]

Answer 1

I have no idea from where to start this.

\[\large{ e^{\tan x} - \log \left( \cfrac{1}{\tan x} \right) = \csc x - \sec x }\]

Answer 2

\(\color{blue}{\text{Originally Posted by}}\) @mathslover
\[ \large{ e^{\tan x} - \log \left( \tan (x) \right) ^{-1} = \csc x - \sec x \\

e^{\tan x} + \log \left(\tan x \right) = \csc x - \sec x } \]

is this okay?
\(\color{blue}{\text{End of Quote}}\)

Answer 3

@experimentX @BSwan and @Miracrown

Answer 4

do you want exact solution or approximate solution?

Answer 5

Not sure, how will you do to solve this for approximate solution. If you think that finding approximate solution will be okay then I would love to see it.

[I saw this question on Facebook, and posted it here]

Answer 6

there is a well known method on numerical analysis.
http://en.wikipedia.org/wiki/Newton's_method

Answer 7

wolf can't find it ... "standard computation time exceeded"
http://www.wolframalpha.com/input/?i=solve++e%5Etan%28x%29+%2B+sec%28x%29+%3D+log%28cot%28x%29%29+%2B+csc%28x%29

Answer 8

try to find a,b
s,t
f(a)=-ve
f(b)=+ve
then every thing is well be good using any approximation

Answer 9

Well, someone commented there that is it :

\(\large e^{\tan x + \sec x} = \log (\cot x + \csc x) \)

I'm not sure whether this is the question or the above one. I'm sorry.

Answer 10

I never studied Newton's Method. I will have to give some time to it though.

But, @experimentX and @BSwan , do you both think that this will be the best method to solve such questions?

Answer 11

the newton's method is numerical method, if you can avoid it, avoid it.

Answer 12

there is no best way :)
if u can solve exactly then its the best way

Answer 13

Okay. Well,

Let us just suppose that the question was :

\[\large e^{\tan x + \sec x} = \log (\cot x + \csc x ) \]

Any particular ideas, on how to solve this?

Answer 14

im thinking :O
i would chick domain/range for both

Answer 15

simpification yields
e^((sin(x) + 1)/cos(x)) = log((cos(x) + 1)/sin(x))
e^(f(x)) = log(f(pi/2-x))

Answer 16

Oh... nice

Answer 17

the equality occurs when f(x) = -infinity and f(pi/2 - x) = 1

Answer 18

I have a question, what level do you think the question belongs to?

Answer 19

i honestly don't know ... mathematica can't find it's numerical solution.

Answer 20

That one is for the assumed quest. right? (second quest.)

Answer 21

huh?

Answer 22

\(e^{\tan x + \sec x} = \log (\cot x + \csc x) \)

The statement you gave above was for this question? ^

Answer 23

yep

Answer 24

Okay. Though, I'm not getting it completely , but I will again refer to Newton's Method as this seems to be much higher for me, and may be, it would help me in future if I learn Newton's Method.

Thanks @experimentX .

Answer 25

sure ... there seems one solution near at x=0
http://www.wolframalpha.com/input/?i=Plot%5BE%5E%28Tan%5Bx%5D+%2B+Sec%5Bx%5D%29+-+Log%5BCot%5Bx%5D+%2B+Csc%5Bx%5D%5D%2C+%7Bx%2C+-10%2C+10%7D%5D

Answer 26

Oh...Okay! :(

Answer 27

http://www.wolframalpha.com/input/?i=Plot%5BE%5E%28Tan%5Bx%5D%29%2BSec%5Bx%5D+-+Log%5BCot%5Bx%5D%5D-+Csc%5Bx%5D%2C+%7Bx%2C+-10%2C+10%7D%5D

Answer 28

there is one solution very close to 0.5
http://www.wolframalpha.com/input/?i=Plot%5BE%5E%28Tan%5Bx%5D%29+%2B+Sec%5Bx%5D+-+Log%5BCot%5Bx%5D%5D+-+Csc%5Bx%5D%2C+%7Bx%2C+0%2C+1%7D%5D

Answer 29

Yeah... seems to be like 0.49 or 0.48.

Answer 30

hmm ... don't be deceived, it can be irrational as well.

Answer 31

Okay! I think, I am getting now..

Answer 32

ok ok ... i don't feel like doing it right now lol.
I think numerical method is the best to do it right now.
consider let the expression be f(x) = g(x)

then let y = f(x) - g(x) and use newton's method. and best of luck :)
gotta go ... see you later.

Answer 33

Sure. Thanks!