How do you solve ${sec}^{2} (x) - sec (x) = 2$ $sec^2(x) - sec(x) = 2$ ?

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Answer 1 · 2015-12-04T10:35:29

$x = π + 2 k π$ $x= \pi + 2k\pi$

$x = \frac{π}{3} + 2 k π$ $x = \pi/3 + 2k\pi$

$x = - \frac{π}{3} + 2 k π$ $x= -pi\/3 + 2k\pi$

Since $sec (x) = \frac{1}{cos (x)}$ $sec(x)=1/cos(x)$ , the expression becomes

$\frac{1}{{cos}^{2} (x)} - \frac{1}{cos (x)} = 2$ $1/cos^2(x) - 1/cos(x) = 2$

Assuming $cos (x) \neq 0$ $cos(x)\ne 0$ , we can multiply everything by ${cos}^{2} (x)$ $cos^2(x)$ :

$1 - cos (x) = 2 {cos}^{2} (x)$ $1-cos(x) = 2cos^2(x)$ .

Rearrange:

$2 {cos}^{2} (x) + cos (x) - 1 = 0$ $2cos^2(x)+cos(x)-1=0$ .

Set $t = cos (x)$ $t=cos(x)$ :

$2 t^{2} + t - 1 = 0$ $2t^2+t-1=0$

Solve as usual with the discriminant formula:

$t = - 1$ $t=-1$ , $t = \frac{1}{2}$ $t=1/2$

Convert the solutions:

$cos (x) = - 1 \Leftrightarrow x = π + 2 k π$ $cos(x)=-1 \iff x=\pi+2k\pi$

$cos (x) = \frac{1}{2} \Leftrightarrow x = \pm \frac{π}{3} + 2 k π$ $cos(x)=1/2 \iff x=\pm\pi/3 +2k\pi$

How do you solve sec^2(x) - sec(x) = 2sec2(x)−sec(x)=2sec^2(x) - sec(x) = 2?