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

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

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Answer 1 · 2018-07-15T17:23:40

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

Explanation:

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

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

Solving this quadratic in $x^{2} \geq 0$ $x^2 >= 0$ ,

$x^{2} = \sqrt{2} + 1 \Rightarrow tan θ = x = \pm \sqrt{\sqrt{2} + 1}$ $x^2 = sqrt 2 + 1 rArr tan theta = x = +- sqrt(sqrt 2 + 1)$

$= \pm 1.5538$ $= +-1.5538$ # ( see graph ), giving

$θ = k π + arctan (\pm \sqrt{\sqrt{2} \pm 1})$ $theta = kpi + arctan ( +-sqrt(sqrt 2+- 1))$ , k = 0, +-1, +-2, +-3, ...#

$= k π \pm {57.235}^{o}, π = 180^{o}$ $= kpi +- 57.235^o, pi = 180^o$
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

$(1 + tan θ) {sec}^{2} θ + 2 {tan}^{2} θ = 0$ $(1 + tan theta ) sec^2theta + 2 tan^2theta = 0$ ,

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

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

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

$=- 70.145^o = -1.2243 rad > - pi/2 = -1.57079...$ 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]}

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

1 Answer

Explanation:

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