# Quod Erat Demonstrandum

## 2012/10/08

### 錯在哪裡系列之求角度

Filed under: mathematics,NSS — johnmayhk @ 12:01 下午

Question

For real numbers $x$, define $f(x+\frac{1}{x})=\frac{x^2}{1+x^2+x^4}$.

Solve $f(\sin\theta)=-\frac{4}{3}$

for $0^o\le \theta \le 360^o$.

Solution

$f(x+\frac{1}{x})$

$=\frac{x^2}{1+x^2+x^4}$

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

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

Hence

$f(u)=\frac{1}{u^2-1}$

Now

$f(\sin\theta)=\frac{1}{\sin^2\theta-1}=-\frac{4}{3}$

$\sin^2\theta=\frac{1}{4}$

$\sin\theta=\pm \frac{1}{2}$

$\therefore \theta=30^o,150^o,210^o,330^o$

## 6 則迴響 »

1. x cannot be zero, so the subject change has problem.
Is it correct, Ng sir?

迴響 由 Little Maths — 2012/10/11 @ 12:47 下午 | 回應

• Thank you for trying!

As you mentioned, the change is a problem.

Precisely, the absolute value of $x+\frac{1}{x}$ is not less than 2 for any non-zero real numbers x. (Why?)

Hence the substitution of $x+\frac{1}{x}=\sin\theta$ is invalid!

迴響 由 johnmayhk — 2012/10/11 @ 4:20 下午 | 回應

• (sqrt(x)-1/sqrt(x))^2 > or =0
x-2+1/x > or = 0
x+1/x not less than2

迴響 由 jeffrey yan — 2012/10/12 @ 6:27 下午

2. My try:
for real $x$, $\frac{x^2}{1+x^2+x^4}$ cannot be negative. So no definition of $f$ is given when $f$ is negative.
Therefore, the equation cannot be solved.

迴響 由 mm100100 — 2012/10/11 @ 7:21 下午 | 回應

3. Good jeffrey,

But be careful, you need to consider the case for the value of x being less than zero.

As I mentioned, it should be the ABSOLUTE value of (x+1/x).

迴響 由 johnmayhk — 2012/10/13 @ 1:07 下午 | 回應