$\displaystyle cosx + \frac{1}{cos^2x} - sinx - \frac{1}{sin^2x} = 0$

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

I hate these kind of equations. :(

Printable View

- Jan 11th 2008, 10:26 PMSpecTricky equation
$\displaystyle cosx + \frac{1}{cos^2x} - sinx - \frac{1}{sin^2x} = 0$

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

I hate these kind of equations. :( - Jan 12th 2008, 12:05 AMPeritus
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, 02:02 AMmr fantastic
- Jan 12th 2008, 04:48 AMSpec
What about the other root(s)?

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

EDIT: And the function is $\displaystyle sinx + cosx + tanx + cotx$ - Jan 12th 2008, 06:39 AMPeritus
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, 07:20 AMJaneBennet
Let $\displaystyle X=\cos{x},\ Y=\sin{x}$. Then you need to solve the following pair of equations:

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

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

By symmetry, $\displaystyle X=Y$. - Jan 12th 2008, 08:22 AMSean12345
- Jan 12th 2008, 03:47 PMSpec
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.