Find x , where 0<x<360 , which satisfy

$\displaystyle \sec x +\tan x=4$

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

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

But how come it doesn't work when

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

is it because -1 is not positive ? Is it one of the conditions ?

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

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

But how come it doesn't work when

i put $\displaystyle \sin x-4\cos x=$ -1
I do not understand what you are saying;
if you multiply both sides of this
$\displaystyle 4\cos x-\sin x=1.0019...$

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

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

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

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

$\displaystyle \sin x-4\cos x=$ -1.0019...
sorry bout that Aidan , edited .