# Tricky equation

• Jan 11th 2008, 11:26 PM
Spec
Tricky equation
$cosx + \frac{1}{cos^2x} - sinx - \frac{1}{sin^2x} = 0$

I can see one solution straight away ( $\pi/4 + n\pi$), but I can't actually solve the equation.

I hate these kind of equations. :(
• Jan 12th 2008, 01:05 AM
Peritus
to solve this equation we'll use the following identities:

1. 1+(tan(x))^2 = 1/(cosx)^2
2. 1+(cot(x))^2 = 1/(sinx)^2

after applying them to the given equation we get:

(tan(x))^2 + cos(x) - sin(x) - (cot(x))^2 = 0

<=> [(sin(x))^4 - (cos(x))^4]/[(sin(x)*cos(x))^2] + cos(x) - sin(x) = 0
<=> (sin(x)^2 + cos(x)^2)*(sin(x)^2 - cos(x)^2) + (cos(x)^3)*(sin(x)^2) - (sin(x)^3)*(cos(x)^2) = 0

this equation holds only in those places where sin(2x)^2 doesn't vanish -> x != (pi/2) * k

<=> (sin(x) + cos(x))*(sin(x) - cos(x)) + (cos(x)^2)*(sin(x)^2)*[cos(x) - sin(x)] = 0
<=> (sin(x) - cos(x))*[sin(x) + cos(x) - (cos(x)^2)*(sin(x)^2)] = 0

sin(x) = cos(x)

as you've already mentioned one solution to the equation is pi/4 + pi*k...
• Jan 12th 2008, 03:02 AM
mr fantastic
Quote:

Originally Posted by Peritus
to solve this equation we'll use the following identities:

1. 1+(tan(x))^2 = 1/(cosx)^2
2. 1+(cot(x))^2 = 1/(sinx)^2

after applying them to the given equation we get:

(tan(x))^2 + cos(x) - sin(x) - (cot(x))^2 = 0

<=> [(sin(x))^4 - (cos(x))^4]/[(sin(x)*cos(x))^2] + cos(x) - sin(x) = 0
<=> (sin(x)^2 + cos(x)^2)*(sin(x)^2 - cos(x)^2) + (cos(x)^3)*(sin(x)^2) - (sin(x)^3)*(cos(x)^2) = 0

this equation holds only in those places where sin(2x)^2 doesn't vanish -> x != (pi/2) * k

<=> (sin(x) + cos(x))*(sin(x) - cos(x)) + (cos(x)^2)*(sin(x)^2)*[cos(x) - sin(x)] = 0
<=> (sin(x) - cos(x))*[sin(x) + cos(x) - (cos(x)^2)*(sin(x)^2)] = 0

sin(x) = cos(x)

as you've already mentioned one solution to the equation is pi/4 + pi*k...

Indeed. This is where I got to as well. The real trick is to solve sin(x) + cos(x) - (cos(x)^2)*(sin(x)^2) = 0. At best, it looks like getting exact solutions will be rather difficult. At worst, well .......
• Jan 12th 2008, 05:48 AM
Spec

What I'm really supposed to do is to show that the function is monotonic on the interval $]0,\frac{\pi}{4}[$, so I thought this would be the easiest approach. I guess it wasn't. :D

EDIT: And the function is $sinx + cosx + tanx + cotx$
• Jan 12th 2008, 07:39 AM
Peritus
Are you allowed to use computer software?
If you do, you can use matlab to find the roots numerically (gradient descent, Newton's method, Picard's iterative process....) to show that the derivative has no roots in the interval [0, pi/4], and thus proving that the function is indeed monotonic (it is continues and well defined in this interval). Alternatively you can follow a more simple approach by defining a vector of sufficiently large resolution in the given interval and and defferentiating the function in this interval (diff), the indexes where the function is within some empirical range from 0 are the extrema points.
• Jan 12th 2008, 08:20 AM
JaneBennet
Let $X=\cos{x},\ Y=\sin{x}$. Then you need to solve the following pair of equations:

$\color{white}.\quad.$ $X+\frac{1}{X^2}=Y+\frac{1}{Y^2}$

$\color{white}.\quad.$ $X^2+Y^2=1$

By symmetry, $X=Y$.
• Jan 12th 2008, 09:22 AM
Sean12345
Quote:

Originally Posted by JaneBennet
Let $X=\cos{x},\ Y=\sin{x}$. Then you need to solve the following pair of equations:

$\color{white}.\quad.$ $X+\frac{1}{X^2}=Y+\frac{1}{Y^2}$

$\color{white}.\quad.$ $X^2+Y^2=1$

By symmetry, $X=Y$.

Nice method.
• Jan 12th 2008, 04:47 PM
Spec
Still, that's only one of the two solutions. How do I get the other (plot it and you'll see that there're two types of roots)?

Also, I'm not quite sure this is how I'm supposed to solve the original problem (showing that the function is monotonic on the specified interval). I think I'm supposed to use the Mean Value Theorem, but I can't really find any use for it if I don't know what the derivative is.