An logarithmic problem and an trigonometrical problem

**λιεҗąиđ€ŗ** · March 1st 2008, 06:19 AM

Hello

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

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

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

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Ok the first one:

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

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

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

$x^2 = 2 - x$

$x^2 +x - 2 = 0$

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

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

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

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

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

A check of these answers seems to give good results.

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2

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

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

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

$1 = -tan(x)$

$-1 = tan(x)$

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

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

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

Thanks for replies

Edit: Is the period exresssion for 2 correct? I'm not too familiar with radians.

**Soroban** · March 1st 2008, 07:27 AM

Hello, λιεҗąиđ€ŗ!

A few minor things . . .

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

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

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

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

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

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

But $x = -2$ is extraneous.

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

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

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

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

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

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