1. ## Solve for x

Solve the equation:
$\displaystyle \arcsin(\frac{x}{x-1}) +2 \arctan(\frac{1}{x+1})= \frac{\pi}{2}$

My attempt:
let $\displaystyle \theta = \arcsin(\frac{x}{x-1})$......1
and $\displaystyle \phi =\arctan(\frac{1}{x+1})$.......2
from these:
$\displaystyle \sin\theta=\frac{x}{x-1}$
$\displaystyle \tan\phi=\frac{1}{x+1}$
substitute 1 and 2 into the equation:
$\displaystyle \theta + 2 \phi = \frac{\pi}{2}$
$\displaystyle \sin (\theta+2\phi) =\sin \frac{\pi}{2}$
$\displaystyle \sin\theta\cos2\phi + \cos\theta\sin2\phi = 1$
substitute $\displaystyle \sin\theta$ and use the identity:
$\displaystyle \sin2\phi=\frac{2t}{1+t^2}$ and $\displaystyle \cos2\phi=\frac{1-t^2}{1+t^2}$ where $\displaystyle t=\tan\phi$
$\displaystyle (\sin\theta)(\frac{1-t^2}{1+t^2}) + (\sin\theta)(\frac{2t}{1+t^2})=1$
$\displaystyle (\frac{x}{x-1})(\frac{1-(\frac{1}{x+1})^2}{1+(\frac{1}{x+1})^2}) + (\cos\theta)(\frac{2(\frac{1}{x+1})}{1+(\frac{1}{x +1})^2})=1$
my problem is the $\displaystyle \cos\theta$, I tried using pythagoras' theorem to find the the length of the adjacent side the then use that to determine the cosine, but I can't work it out.
thanks for any help.

2. ## Trig equation

Hello arze
Originally Posted by arze
Solve the equation:
$\displaystyle \arcsin(\frac{x}{x-1}) +2 \arctan(\frac{1}{x+1})= \frac{\pi}{2}$

My attempt:
let $\displaystyle \theta = \arcsin(\frac{x}{x-1})$......1
and $\displaystyle \phi =\arctan(\frac{1}{x+1})$.......2
from these:
$\displaystyle \sin\theta=\frac{x}{x-1}$
$\displaystyle \tan\phi=\frac{1}{x+1}$
substitute 1 and 2 into the equation:
$\displaystyle \theta + 2 \phi = \frac{\pi}{2}$ OK up to here - see below.
$\displaystyle \sin (\theta+2\phi) =\sin \frac{\pi}{2}$
$\displaystyle \sin\theta\cos2\phi + \cos\theta\sin2\phi = 1$
substitute $\displaystyle \sin\theta$ and use the identity:
$\displaystyle \sin2\phi=\frac{2t}{1+t^2}$ and $\displaystyle \cos2\phi=\frac{1-t^2}{1+t^2}$ where $\displaystyle t=\tan\phi$
$\displaystyle (\sin\theta)(\frac{1-t^2}{1+t^2}) + (\sin\theta)(\frac{2t}{1+t^2})=1$
$\displaystyle (\frac{x}{x-1})(\frac{1-(\frac{1}{x+1})^2}{1+(\frac{1}{x+1})^2}) + (\cos\theta)(\frac{2(\frac{1}{x+1})}{1+(\frac{1}{x +1})^2})=1$
my problem is the $\displaystyle \cos\theta$, I tried using pythagoras' theorem to find the the length of the adjacent side the then use that to determine the cosine, but I can't work it out.
thanks for any help.
From the point that I indicated:

$\displaystyle \theta = \tfrac{\pi}{2} - 2\phi$

$\displaystyle \Rightarrow \sin\theta = \sin(\tfrac{\pi}{2} - 2\phi) = \cos2\phi$

$\displaystyle \Rightarrow \frac{x}{x-1}=\frac{1-t^2}{1+t^2}$

$\displaystyle \Rightarrow \frac{x}{x-1}=\frac{1-\frac{1}{(x+1)^2}}{1+\frac{1}{(x+1)^2}}$

$\displaystyle = \frac{(x^2+2x+1)-1}{(x^2+2x+1)+1}$

$\displaystyle =\frac{x(x+2)}{x^2+2x+2}$

$\displaystyle \Rightarrow x = 0$ or $\displaystyle \frac{1}{x-1}=\frac{x+2}{x^2+2x+2}$

$\displaystyle \Rightarrow x^2+2x+2 = (x-1)(x+2)$

$\displaystyle \Rightarrow x = -4$

So the solutions are $\displaystyle x = 0, -4$

3. How did you get x=0? i don't understand how you arrived at that solution.
also the answer is supposed to be only 0, but i think i know why.

4. hi
the equation $\displaystyle x^2+2x+2$ has no real roots.
to solve $\displaystyle \frac{x(x+2)}{x^2+2x+2}$ only $\displaystyle x(x+2)$ must be equal to zero,so the solutions are $\displaystyle \mathbb{S}=\left \{ 0,-2 \right \}$

5. Hello, arze!

There must be a typo . . .

Solve the equation:

. . $\displaystyle \arcsin\underbrace{\left(\frac{x}{x-1}\right)} +2 \arctan\left(\frac{1}{x+1}\right) \;=\; \frac{\pi}{2}$
. . . . $\displaystyle {\color{blue}^{\text{This is greater than 1}}}$

6. i think $\displaystyle x$ must be a negative integer or zero
$\displaystyle Arcsin : [-1,1] \mapsto \left [ \frac{-\pi}{2},\frac{\pi}{2} \right ]$

7. Hello arze
Originally Posted by arze
How did you get x=0? i don't understand how you arrived at that solution.
also the answer is supposed to be only 0, but i think i know why.
Because $\displaystyle x$ is a factor of both sides of the equation $\displaystyle \frac{x}{x-1}=\frac{x(x+2)}{x^2+2x+2}$

I'm not sure what the problem is with x = -4. I'll think about it!

8. i see! thanks! now the only problem i have left is the -4 value

9. Originally Posted by Soroban
Hello, arze!

There must be a typo . . .
i copied it exactly as it appears in the book, so i had concluded that x is 0 or negative

10. the only solution for x is x = 0; the value x = -4 is not possible.

11. Hello arze
Originally Posted by arze
i see! thanks! now the only problem i have left is the -4 value
The problem of extra 'solutions' is always likely to occur when we square things as part of solving an equation, because squaring a real number always 'hides' a possible minus sign by giving a positive value, whatever the sign of the original number.

It's pretty obvious when you do something like this, for instance:

$\displaystyle 2x-3 = x+4$

$\displaystyle \Rightarrow (2x-3)^2=(x+4)^2$

$\displaystyle \Rightarrow 4x^2 -12x+9 =x^2+8x+16$

$\displaystyle \Rightarrow 3x^2-20x-7=0$

$\displaystyle \Rightarrow (3x+1)(x-7)=0$

$\displaystyle \Rightarrow x=-\tfrac13$ or $\displaystyle x=7$

All of which is correct, but that doesn't mean that you can reverse the implication arrows and say that both values of $\displaystyle x$ are valid solutions to the original equation.

$\displaystyle x=7$ is OK. (This gives $\displaystyle LHS = 11 = RHS$.)

But $\displaystyle x = -\tfrac13$ isn't. This gives $\displaystyle LHS = -\tfrac{11}{3}$, and $\displaystyle RHS = +\tfrac{11}{3}$.

So, whereas $\displaystyle x =-4$ is a valid solution to the equation

$\displaystyle \frac{x}{x-1}=\frac{x(x+2)}{x^2+2x+2}$

it doesn't necessarily mean that it's a solution to the original equation. $\displaystyle x=-4$ gives $\displaystyle \sin\theta = \tfrac45$ and $\displaystyle \tan\phi = -\tfrac13$, whereas you'll find you need $\displaystyle \sin\theta = \tfrac45,\,\tan\phi = +\tfrac13$ to satisfy the original equation.

The problem arises when we use $\displaystyle \cos2\phi=\frac{1-t^2}{1+t^2}$, because this has the same value, $\displaystyle \tfrac45$, whether $\displaystyle t = \tfrac13$ or $\displaystyle t=-\tfrac13$.

So, we all need to take care with signs when we take squares and square roots!