solving the equation

**masiboy** · April 20th 2009, 12:09 AM

Hi there,

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

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

thanks

**red_dog** · April 20th 2009, 01:03 AM

$\frac{2\sin^2x}{\cos^2x}+\frac{2}{\cos^2x}-\frac{1}{\cos x}=12\Rightarrow 12\cos^2x+\cos x-2\sin^2x-2=0$

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

Can you do that?

**masiboy** · April 20th 2009, 01:41 AM

Originally Posted by red_dog

$\frac{2\sin^2x}{\cos^2x}+\frac{2}{\cos^2x}-\frac{1}{\cos x}=12\Rightarrow 12\cos^2x+\cos x-2\sin^2x-2=0$

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

Can you do that?

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

**mr fantastic** · April 20th 2009, 02:50 AM

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 $t = \cos x$ and solve the resulting quadratic equation for t. Then, once you have found the value(s) of t, substitute back into $t = \cos x$ and solve for x.

**Soroban** · April 20th 2009, 04:20 AM

Hello, masiboy!

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

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

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

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

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

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

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

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