# trigo and log equation

• Jun 8th 2009, 12:38 PM
dhiab
trigo and log equation
• Jun 8th 2009, 01:17 PM
Amer
Quote:

Originally Posted by dhiab

I will try

let $t=\pi (ln(x^2))$

[tex]sint + cost = 1 [tex]

$2sin(t)cos(t) + 1 - 2sin^2(t) = 1$

$2sin(t)(cos(t) - sin(t)) = 0$

$sint=0........ln(x^2)=0.....x=\pm 1$

$sint=cost....t=\frac{\pi}{4} + 2\pi n .....$

$\pi ln(x^2)=\pi ( \frac{1}{4} + 2n )$

$ln(x^2)= \frac{1}{4} + 2n$

$x^2 = e^{.75 + 2n }$

$x= e^{.375 + n }$
• Jun 9th 2009, 07:22 AM
dhiab
Quote:

Originally Posted by Amer
I will try

let $t=\pi (ln(x^2))$

[tex]sint + cost = 1 [tex]

$2sin(t)cos(t) + 1 - 2sin^2(t) = 1$

$2sin(t)(cos(t) - sin(t)) = 0$

$sint=0........ln(x^2)=0.....x=\pm 1$

$sint=cost....t=\frac{\pi}{4} + 2\pi n .....$

$\pi ln(x^2)=\pi ( \frac{1}{4} + 2n )$

$ln(x^2)= \frac{1}{4} + 2n$

$x^2 = e^{.75 + 2n }$

$x= e^{.375 + n }$

Hello Amer
here is expression just : sint = 2sin(t/2)cos(t/2) and :
cost = 1-2sin2 (t/2)
TANK YOU(Clapping)
• Jun 9th 2009, 07:53 AM
pankaj
Quote:

Originally Posted by dhiab

Divide by $\sqrt{2}$ on both sides

$\sin \frac{\pi}{4} \sin \pi ln(x^2)+\cos \frac{\pi}{4} \cos \pi ln(x^2)=\frac{1}{\sqrt{2}}$

$
\cos(\frac{\pi}{4}+\pi ln(x^2))=\cos\frac{\pi}{4}
$

$
\frac{\pi}{4}+\pi ln(x^2)=2n\pi \pm \frac{\pi}{4}
$

$ln(x^2)=2n$; $x^2=e^{2n}$; $x=e^{n}$

$ln(x^2)=\frac{1}{2}-2n$; $x^2=e^{\frac{1}{2}-2n}$; $x=e^{\frac{1}{4}-n}$

'n' being any integer
• Jun 9th 2009, 08:07 AM
dhiab
Quote:

Originally Posted by pankaj
Divide by $\sqrt{2}$ on both sides

$\sin \frac{\pi}{4} \sin \pi ln(x^2)+\cos \frac{\pi}{4} \cos \pi ln(x^2)=\frac{1}{\sqrt{2}}$

$
\cos(\frac{\pi}{4}+\pi ln(x^2))=\cos\frac{\pi}{4}
$

$
\frac{\pi}{4}+\pi ln(x^2)=2n\pi \pm \frac{\pi}{4}
$

$ln(x^2)=-2n$; $x^2=e^{-2n}$; $x=e^{-n}$

$ln(x^2)=\frac{1}{2}-2n$; $x^2=e^{\frac{1}{2}-2n}$; $x=e^{\frac{1}{4}-n}$

'n' being any integer

Hello : i' cannot understing this resolution : http://www.mathhelpforum.com/math-he...9b7e2e94-1.gif(Angry)
Thank you
• Jun 9th 2009, 08:30 AM
pankaj
I made a mistake.I have edited my post.