# solving the equation

• Apr 20th 2009, 12:09 AM
masiboy
solving the equation
Hi there,

it would be much appreciated if any one can help me solve this equation

2tan^2x+2sec^2x-secx =12

thanks
• Apr 20th 2009, 01:03 AM
red_dog
$\displaystyle \frac{2\sin^2x}{\cos^2x}+\frac{2}{\cos^2x}-\frac{1}{\cos x}=12\Rightarrow 12\cos^2x+\cos x-2\sin^2x-2=0$

Replace $\displaystyle \sin^2x=1-\cos^2x$, then note $\displaystyle \cos x=t$ and solve the quadratic.

Can you do that?
• Apr 20th 2009, 01:41 AM
masiboy
Quote:

Originally Posted by red_dog
$\displaystyle \frac{2\sin^2x}{\cos^2x}+\frac{2}{\cos^2x}-\frac{1}{\cos x}=12\Rightarrow 12\cos^2x+\cos x-2\sin^2x-2=0$

Replace $\displaystyle \sin^2x=1-\cos^2x$, then note $\displaystyle \cos x=t$ and solve the quadratic.

Can you do that?

sorry but i dont know wat u mean when u say ''note cos x =t
• Apr 20th 2009, 02:50 AM
mr fantastic
Quote:

Originally Posted by masiboy
sorry but i dont know wat u mean when u say ''note cos x =t

red_dog means make the substitution $\displaystyle t = \cos x$ and solve the resulting quadratic equation for t. Then, once you have found the value(s) of t, substitute back into $\displaystyle t = \cos x$ and solve for x.
• Apr 20th 2009, 04:20 AM
Soroban
Hello, masiboy!

I'll assume that the answers are on $\displaystyle [0^o,\,360^o]$

Quote:

$\displaystyle 2\tan^2\!x+2\sec^2\!x-\sec x \:=\:12$

Since $\displaystyle \tan^2\!x \:=\:\sec^2\!x - 1$, the equation becomes: .$\displaystyle 2(\sec^2\!x - 1) + 2\sec^2\!x - \sec x \:=\:12$

. . which simplifies to: .$\displaystyle 4\sec^2\!x - \sec x - 14 \:=\:0$

. . and factors: .$\displaystyle (\sec x - 2)(4\sec x + 7) \:=\:0$

We have: .$\displaystyle \sec x -2 \:=\:0 \quad\Rightarrow\quad \sec x \:=\:2\quad\Rightarrow\quad x \:=\:60^o,\:300^o$

. . and: .$\displaystyle 4\sec x + 7 \:=\:\quad\Rightarrow\quad \sec x \:=\:-\frac{7}{4}\quad\Rightarrow\quad x \;\approx\;124.85^o,\:235.15^o$