# Find the number of solutions of equation: (1-tan θ)(1+tan θ) sec²θ+2^(tan²θ)=0 when θ∈(-π/2,π/2)?

Jul 15, 2018

$\theta = k \pi + \arctan \left(\pm \sqrt{\sqrt{2} + 1}\right)$,
$k = 0 , \pm 1 , \pm 2 , \pm 3 , . . = k \pi \pm {57.235}^{o} , \pi = {180}^{o}$
..

#### Explanation:

Use ${\sec}^{2} \theta = 1 + {\tan}^{2} \theta$ and $x = \tan \theta .$

The equation becomes the cubic ${x}^{4} - 2 {x}^{2} - 1 = 0$

Solving this quadratic in ${x}^{2} \ge 0$,

${x}^{2} = \sqrt{2} + 1 \Rightarrow \tan \theta = x = \pm \sqrt{\sqrt{2} + 1}$

$= \pm 1.5538$  ( see graph ), giving

$\theta = k \pi + \arctan \left(\pm \sqrt{\sqrt{2} \pm 1}\right)$, k = 0, +-1, +-2, +-3, ...

$= k \pi \pm {57.235}^{o} , \pi = {180}^{o}$
graph{y - x^4 + 2 x^2 +1 = 0[-2 2 -0.01 0.01]}

Sometimes, oversight miss produces good results !

What follows is the solution for mistaken equation

$\left(1 + \tan \theta\right) {\sec}^{2} \theta + 2 {\tan}^{2} \theta = 0$,

$\Rightarrow {x}^{3} + 3 {x}^{2} + x + 1 = 0$

Graph locates only one real root x = - 2.75, nearly.

Astute scaling nearby, approximates x to 5-sd $- 2.7693$.
$\theta = {\tan}^{- 1} x = \tan \left(- 1\right) \left(- 2.7693\right)$

$= - {70.145}^{o} = - 1.2243 r a d > - \frac{\pi}{2} = - 1.57079 \ldots$ rad.

graph{y-x^3-3x^2-x-1=0}
graph{y-x^3-3x^2-x-1=0[-2.7695 -2.769 -.01 .01]}