The general solution of [tex]\tan bx = 2[/tex] and [tex]x = 0.3[/tex] is [tex]x = 0.095\pi \mp 0.271\pi\cdot i[/tex], [tex]\forall \,i\in \mathbb{N}_{O}[/tex].
From Trigonometry we remember that Tangent is a Transcendental Function that is positive both in 1st and 3rd Quadrants and have a periodicity of [tex]\pi[/tex] radians. The procedure consists in using concepts of Direct and Inverse Trigonometric Functions as well as characteristics related to the behavior of the tangent function in order to derive a General Formula for every value of [tex]x[/tex], measured in radians.
First, we solve the following system of equations for [tex]b[/tex]:
[tex]\tan bx = 2[/tex] (1)
[tex]x = 0.3[/tex] (2)
Please notice that angles are measured in radians.
(2) in (1):
[tex]\tan 0.3b = 2[/tex]
[tex]0.3\cdot b = \tan^{-1} 2[/tex]
[tex]b = \frac{10}{3}\cdot \tan^{-1}2[/tex]
[tex]b\approx 3.690[/tex]
Under the assumption of periodicity, we know that:
[tex]y = \tan bx[/tex]
[tex]b\cdot x \pm \pi\cdot i = \tan^{-1} y[/tex], [tex]\forall \,i\in \mathbb{N}_{O}[/tex]
[tex]b\cdot x = \tan^{-1}y \mp \pi\cdot i[/tex]
[tex]x = \frac{\tan^{-1}y \mp \pi\cdot i}{b}[/tex]
If we know that [tex]y = 2[/tex] and [tex]b \approx 3.690[/tex], then the general solution of this trigonometric function is:
[tex]x = \frac{0.352\pi \mp \pi\cdot i}{3.690}[/tex], [tex]\forall \,i\in \mathbb{N}_{O}[/tex]
[tex]x = 0.095\pi \mp 0.271\pi\cdot i[/tex], [tex]\forall \,i\in \mathbb{N}_{O}[/tex]
The general solution of [tex]\tan bx = 2[/tex] and [tex]x = 0.3[/tex] is [tex]x = 0.095\pi \mp 0.271\pi\cdot i[/tex], [tex]\forall \,i\in \mathbb{N}_{O}[/tex].
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