# An logarithmic problem and an trigonometrical problem

• Mar 1st 2008, 06:19 AM
λιεҗąиđ€ŗ
An logarithmic problem and an trigonometrical problem
Hello :)

I'd like an check on my answers for these two equations:

1 $\displaystyle 2ln(x) = ln(2 - x) + ln(1)$

2 $\displaystyle cos(-x) = sin (-x)$ | $\displaystyle -2\pi \leq x \leq 2\pi$
Note: We're working in radians here

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Ok the first one:

$\displaystyle 2ln(x) = ln(2 - x) + ln(1)$

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

$\displaystyle ln(x^2) = ln(2 - x)$

$\displaystyle x^2 = 2 - x$

$\displaystyle x^2 +x - 2 = 0$

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

$\displaystyle x = - \frac{1}{2} \pm \sqrt{\frac{9}{4}}$

$\displaystyle x = - \frac{1}{2} \pm \frac{3}{2}$

$\displaystyle x_{1} = \frac{2}{2}$

$\displaystyle x_{2} = -\frac{4}{2}$

A check of these answers seems to give good results.

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2

$\displaystyle cos(-x) = sin (-x)$

$\displaystyle 1 = \frac{sin (-x)}{cos(-x)}$

$\displaystyle 1 = - \frac{sin (x)}{cos(x)}$

$\displaystyle 1 = -tan(x)$

$\displaystyle -1 = tan(x)$

$\displaystyle x = tan^{-1} (-1)$

$\displaystyle x = - \frac{\pi}{4}$

$\displaystyle x = - \frac{\pi}{4} + n \cdot \pi$

Thanks for replies :)

Edit: Is the period exresssion for 2 correct? I'm not too familiar with radians.
• Mar 1st 2008, 07:27 AM
Soroban
Hello, λιεҗąиđ€ŗ!

A few minor things . . .

Quote:

I'd like an check on my answers for these two equations:

$\displaystyle 1)\;\;2\ln(x) \:= \:\ln(2 - x) + \ln(1)$

My work: .$\displaystyle 2\ln(x) \:= \:\ln(2 - x) + \ln(1) \quad\Rightarrow\quad \ln(x^2) \:= \:\ln(2 - x) + 0$

. . $\displaystyle \ln(x^2) \:= \:\ln(2 - x) \quad\Rightarrow\quad x^2 \:= \:2 - x \quad\Rightarrow\quad x^2 +x - 2 \:= \:0$ . . . . Right!

Factor: .$\displaystyle (x-1)(x+2) \:=\:0$

Hence: .$\displaystyle x \:=\:1,-2$

But $\displaystyle x = -2$ is extraneous.

Solution: .$\displaystyle x \:= \:1$

Quote:

$\displaystyle 2)\;\;\cos(-x) \:= \sin(-x),\quad -2\pi \leq x \leq 2\pi$

My work: .$\displaystyle \cos(-x) \:= \:\sin (-x)\quad\Rightarrow\quad 1 \:= \:\frac{\sin (-x)}{\cos(-x)} \quad\Rightarrow\quad 1 \:=\: -\frac{\sin (x)}{\cos(x)}$

. . $\displaystyle 1 \:= \:-\tan(x) \quad\Rightarrow\quad -1 \:=\: \tan(x)\quad\Rightarrow\quad x \:= \:\tan^{-1}(-1)$ . . . . Correct!

$\displaystyle x \;=\;-\frac{5\pi}{4},\;-\frac{\pi}{4},\;\frac{3\pi}{4},\;\frac{7\pi}{4}$ . . . . on the interval they gave us.