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

#theta = kpi + arctan ( +-sqrt(sqrt 2 + 1))#,
# k = 0, +-1, +-2, +-3, ..= kpi +- 57.235^o, pi = 180^o#
..

Use #sec^2theta= 1 + tan^2theta# and #x = tan theta.#

The equation becomes the cubic #x^4 - 2 x^2 - 1 = 0#

Solving this quadratic in #x^2 >= 0#,

#x^2 = sqrt 2 + 1 rArr tan theta = x = +- sqrt(sqrt 2 + 1)#

#= +-1.5538# # ( see graph ), giving

#theta = kpi + arctan ( +-sqrt(sqrt 2+- 1))#, k = 0, +-1, +-2, +-3, ...#

#= 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 theta ) sec^2theta + 2 tan^2theta = 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#.
#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]}