Question

The smallest positive solution of tan bx = 2 is x = 0.3. Determine the general solution of tan bx = 2.

Answer 1

The general solution of $\tan bx = 2$ and $x = 0.3$ is $x = 0.095\pi \mp 0.271\pi\cdot i$, $\forall \,i\in \mathbb{N}_{O}$.

From Trigonometry we remember that Tangent is a Transcendental Function that is positive both in 1st and 3rd Quadrants and have a periodicity of $\pi$ 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 $x$, measured in radians.

First, we solve the following system of equations for $b$:

$\tan bx = 2$ (1)

$x = 0.3$ (2)

Please notice that angles are measured in radians.

(2) in (1):

$\tan 0.3b = 2$

$0.3\cdot b = \tan^{-1} 2$

$b = \frac{10}{3}\cdot \tan^{-1}2$

$b\approx 3.690$

Under the assumption of periodicity, we know that:

$y = \tan bx$

$b\cdot x \pm \pi\cdot i = \tan^{-1} y$, $\forall \,i\in \mathbb{N}_{O}$

$b\cdot x = \tan^{-1}y \mp \pi\cdot i$

$x = \frac{\tan^{-1}y \mp \pi\cdot i}{b}$

If we know that $y = 2$ and $b \approx 3.690$, then the general solution of this trigonometric function is:

$x = \frac{0.352\pi \mp \pi\cdot i}{3.690}$, $\forall \,i\in \mathbb{N}_{O}$

$x = 0.095\pi \mp 0.271\pi\cdot i$, $\forall \,i\in \mathbb{N}_{O}$

The general solution of $\tan bx = 2$ and $x = 0.3$ is $x = 0.095\pi \mp 0.271\pi\cdot i$, $\forall \,i\in \mathbb{N}_{O}$.

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Answer 2

The general solution of tan(b·x) = 2, given that the smallest positive solution is x = 0.3, is presented as follows;

$\mathbf{ x =0.3 + \dfrac{n \cdot \pi}{b}}$

The given smallest positive solution of tan (b·x) = 2 is x = 0.3

The general solution of tan (b·x + c) = m, is given as follows;

$\mathbf{x = \dfrac{\alpha - c}{b} + n \cdot \dfrac{\pi}{b}}$

α = arctan(m) = x₀

The minimum positive value of the general solution is therefore presented as follows;

$x_{min \ positive} = \dfrac{\alpha - c}{b} + 0 \times \dfrac{\pi}{b} = \dfrac{\alpha - c}{b}$

$\mathbf{x_{min \ positive} = \dfrac{\alpha - c}{b}}$

In the given function, tan (b·x), we therefore, have;

c = 0, m = 2,  $\dfrac{\alpha - c}{b} = 0.3$

The general solution of tan(b·x) = 2, is therefore;

$\mathbf{x =0.3 + \dfrac{n \cdot \pi}{b}}$

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