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Chapter 9: Problem 21

Solve $$ \tan \left(\frac{2 x}{3}\right)=0.7 \quad 0 \leq x \leq 2 \pi $$

Short Answer

Expert verified
Question: Find the values of x within the domain \(0 \leq x \leq 2\pi\) that satisfy the following equation: $$ \tan \left(\frac{2 x}{3}\right)=0.7 $$ Answer: There are two values of x within the given domain that satisfy the equation: \(x_1 \approx 0.9160\) radians, and \(x_2 \approx 3.0577\) radians.

Step by step solution

01

Identify the inverse tangent function

To find the value of \(x\) that will make the tangent function equal to 0.7, we need to find the inverse tangent of 0.7: $$ \theta = \tan^{-1}(0.7) $$
02

Compute the inverse tangent of 0.7

Use a calculator to find the angle whose tangent is 0.7: $$ \theta = \tan^{-1}(0.7) \approx 0.6107 \,\text{radians} $$ Take note that the tangent function has a periodicity of \(\pi\). This means that we need to consider all the angles that are equal to \(\theta\) plus integer multiples of \(\pi\).
03

Determine the range of angles within the given domain

Given that \(0 \leq x \leq 2\pi\), we need to find all the angles \(\theta\) within this domain to which the tangent function is 0.7. Thus, we have: $$ \frac{2x}{3} = \theta + k\pi, \quad k \in \mathbb{Z} $$ To make sure we find all angles that satisfy the condition, we need to consider the largest and smallest possible values for \(x\). In this case, the values will be 0 and \(2\pi\).
04

Solve for x for each value of k within the domain

Substitute the value of \(\theta\) from step 2, and solve the equation from step 3 for \(x\): $$ x = \frac{3}{2}(\theta + k\pi), \quad k \in \mathbb{Z} $$ Now, we need to find all possible values of x that satisfy the condition while respecting the given domain.
05

Identify all possible x-values within the given domain

Plug in different integer values for \(k\), until we find all possible x-values within the domain \(0 \leq x \leq 2\pi\): $$ x = \frac{3}{2}(0.6107 + k\pi) $$ For \(k = 0\): $$ x_1 = \frac{3}{2}(0.6107 + 0\pi) \approx 0.9160 $$ For \(k = 1\): $$ x_2 = \frac{3}{2}(0.6107 + 1\pi) \approx 3.0577 $$ For \(k = 2\) and higher values, x will exceed the boundary of the domain. So, we have two solution in the given domain: \(x_1 \approx 0.9160\) radians, and \(x_2 \approx 3.0577\) radians.

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