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elena55 [62]
3 years ago
7

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

Mathematics
2 answers:
Solnce55 [7]3 years ago
6 0

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 <em>1st</em> and <em>3rd</em> 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}.

For further information, you can see the following outcomes from another users:

brainly.com/question/3056589

brainly.com/question/11526967

Contact [7]3 years ago
5 0

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}}

Learn more about the general solution of a sine function here:

https://brainly.in/question/1549935

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