another solving trigomonetry equation for x

[songoku]

1. The problem statement, all variables and given/known data
Solve:
sin x + cos x + tan x + cot x + sec x + cosec x + 2 = 0


2. Relevant equations
trigonometry


3. The attempt at a solution
The best I can get is:
sin2x cos x + cos2x sin x + sin x + cos x + 3 = 0

Then....:cry:

[ArcanaNoir]

This page has more trig identities than I even knew existed. Maybe you can find something here that will help. http://www.sosmath.com/trig/Trig5/trig5/trig5.html

[songoku]

This page has more trig identities than I even knew existed. Maybe you can find something here that will help. http://www.sosmath.com/trig/Trig5/trig5/trig5.html

I know those identities but I don't know how to use it to solve the problem :redface:

[dextercioby]

Take everything to a common denominator first. Then make \sin x = p to obtain an algebraic eqn in p.

[ArcanaNoir]

Sorry I couldn't be more helpful. I'm not deliberately being unhelpful, I did try to solve the equation, just haven't gotten anywhere on it. :(

[songoku]

Take everything to a common denominator first.
I've done that. I got sin2x cos x + cos2x sin x + sin x + cos x + 3 = 0 as in my first post

Then make \sin x = p to obtain an algebraic eqn in p.
So,
p2 √1-p2 + (1-p2) p + p + √1-p2 + 3 = 0

Can it be solved?

Sorry I couldn't be more helpful. I'm not deliberately being unhelpful, I did try to solve the equation, just haven't gotten anywhere on it. :(
No problem. Thanks for the link :smile:

[dextercioby]

It doesn't like having real solutions...http://www.wolframalpha.com/input/?i=Solve+%28sin+x+cos+x+%2B1%29+%28sin+x+%2Bcos+x% 29+%3D+-3

[ArcanaNoir]

my calculator said:
x=2\cdot \pi \cdot \textrm{constn} (1)-\frac{\pi }{4} ,
x=2\cdot \pi \cdot \textrm{constn} (2)+\frac{3\pi }{4}

I don't see this one working out nicely. >_<

[I like Serena]

Wolfram can solve the original equation directly.
He also rewrites it into a product of trig functions.

The question is how to get there without too much hassle...

[ArcanaNoir]

The cavalry is here! :D

[I like Serena]

Solve:
sin x + cos x + tan x + cot x + sec x + cosec x + 2 = 0

The best I can get is:
sin2x cos x + cos2x sin x + sin x + cos x + 3 = 0

You made a mistake when you derived the second equation.
You forgot to multiply 2 with (sin x cos x).

If you fix that, you can apply for instance dextercioby's method.
If you work it out, you should be able to get a 3rd degree polynomial that you can solve.

[I like Serena]

The cavalry is here! :D

Your attitude towards me seems to have shifted somewhat, since we've first met. ;)
Thanks for the vote of confidence!

[Curious3141]

The only elementary thing you can immediately deduce from that equation is that if theta is a solution, then (pi/2 - theta) is also a solution.

This follows immediately from the fact that replacing x with (pi/2 - x) in the equation changes nothing.

I can't see a shortcut to a rigorous full solution. However, simple inspection immediately yields -pi/4 as a solution, since one needs to find a value where the numerical values of the sine and cosine have to be equal in magnitude and of opposite sign so they cancel out (and the same would apply to the sec and cosec terms), and where the tangent and cotangent are both equal to -1, so they would add to -2 and cancel out the +2 constant term.

Having found -pi/4 as a solution, we known that pi/2 - (-pi/4) = 3pi/4 is also a solution.

One can immediately add the requisite 2k*pi to each of those solutions for the general solution set. Problem is, one can never be sure one is not missing some obscure solution(s) with this sort of slipshod "easy" method.:frown:

[songoku]

You made a mistake when you derived the second equation.
You forgot to multiply 2 with (sin x cos x).

If you fix that, you can apply for instance dextercioby's method.
If you work it out, you should be able to get a 3rd degree polynomial that you can solve.

Ops my bad. The equation should be:
sin2x cos x + cos2x sin x + 1+ sin x + cos x + 2 sin x cos x = 0

let sin x = p
p2√1-p2 + p - p3 + 1 + p + √1-p2 + 2 p √1-p2 = 0

After some works, the equation becomes:
(p + 1) [(√1-p2)(p+1)-p2+p+1)] = 0

To solve [(√1-p2)(p+1)-p2+p+1)] = 0 :
(√1-p2)(p+1) = p2-p-1

squaring both sides and doing some works, I got:
2p4 - p3 - p2 + p = 0
p (2p3 - p2 - p + 1) = 0
p = 0 or 2p3 - p2 - p + 1 = 0

But when I subs p = 0 to (p + 1) [(√1-p2)(p+1)-p2+p+1)] = 0, it is not correct. So it means that p = 0 is extra root?

Then, how to solve 2p3 - p2 - p + 1 = 0 ??

Thanks

[I like Serena]

(√1-p2)(p+1) = p2-p-1

Yep! I'm with you so far.


squaring both sides and doing some works, I got:
2p4 - p3 - p2 + p = 0

I get something else, so I think you made a mistake somewhere...

[songoku]

I get something else, so I think you made a mistake somewhere...

Yeah...:tongue:

Maybe it should be like this:
2p4 - p2 = 0

Am I correct?

[I like Serena]

Yes! That looks much simpler! :)

[songoku]

Yes! That looks much simpler! :)

Then p = 0 is extra root? Because it is not correct when I subs. it into (√1-p2)(p+1) = p2-p-1

[I like Serena]

Yes, p=0 is an extra root that was introduced when you squared both sides of the equation.
If you substitute it, you'll find 1=-1.
When you square both sides, you get 1=1, which is the reason you found it as a solution.

As you can see, you have to verify each solution against the original equation to see if it really is a solution.

[songoku]

Yes, p=0 is an extra root that was introduced when you squared both sides of the equation.
If you substitute it, you'll find 1=-1.
When you square both sides, you get 1=1, which is the reason you found it as a solution.

As you can see, you have to verify each solution against the original equation to see if it really is a solution.

Ok. Thanks a lot for your help :)

[I like Serena]

Thanks!
But then, that's what the cavalry is for. ;)