• Aug 24th 2009, 12:31 AM
thereddevils
Find x , where 0<x<360 , which satisfy

$\sec x +\tan x=4$

$\frac{1+\sin x}{\cos x}=4\Rightarrow 1+\sin x=4\cos x$

putting $4\cos x-\sin x=1$ and continuing , i found that x=61.9 using $r\cos (x+\alpha)=1$ where $r=\sqrt{17}$ and $\alpha=\tan^{-1}\frac{1}{4}$

But how come it doesn't work when

i put $\sin x-4\cos x=$ -1 , doing the same thing .. $r\sin (x-\alpha)=-1$ and i don get the same result , x=61.9 ?

is it because -1 is not positive ? Is it one of the conditions ?
• Aug 24th 2009, 01:18 AM
aidan
Quote:

Originally Posted by thereddevils
...
$\frac{1+\sin x}{\cos x}=4\Rightarrow 1+\sin x=4\cos x$

$4\cos x-\sin x=1$ and continuing , i found that x=61.9

But how come it doesn't work when

i put $\sin x-4\cos x=$ -1

I do not understand what you are saying;
if you multiply both sides of this
$4\cos x-\sin x=1.0019...$

by -1, after rearranging, you will get this:

$\sin x-4\cos x=$ -1.0019...
• Aug 24th 2009, 03:08 AM
thereddevils
Quote:

Originally Posted by aidan
I do not understand what you are saying;
if you multiply both sides of this
$4\cos x-\sin x=1.0019...$

by -1, after rearranging, you will get this:

$\sin x-4\cos x=$ -1.0019...

sorry bout that Aidan , edited .