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Chapter 5: Problem 70

Use inverse functions where needed to find all solutions of the equation in the interval \([0,2 \pi)\). $$\cot ^{2} x-6 \cot x+5=0$$

Short Answer

Expert verified
The solutions of the equation in the interval [0,2蟺) are \(x = \frac{\pi}{4}, \frac{5\pi}{4}, \arctan(0.2), \arctan(0.2) + \pi \).

Step by step solution

01

Convert the given equation in terms of tan(x)

Given the relation \(\cot x = \frac{1}{\tan x}\), the equation \(\cot ^{2} x-6 \cot x+5=0\) can be rewritten as \((\frac{1}{\tan x})^2 - 6*\frac{1}{\tan x} + 5 = 0\). Now, let \(p = \tan x\), therefore the equation becomes: \( \frac{1}{p^{2}} - \frac{6}{p} + 5 = 0\) which is a quadratic in \( \frac{1}{p}\).
02

Solve the Quadratic Equation

To solve the quadratic equation, multiply the entire equation by \(p^2\) to convert into standard form. This gives \(1-6p+5p^2 = 0\), or it can be written as \(5p^2 - 6p + 1 = 0\). Solving this equation for \(p\) we get \( p = \tan x = \frac{6\pm\sqrt{36-4*5*1}}{2*5} = \frac{6\pm\sqrt{16}}{10} = \frac{6\pm4}{10} = 1, \frac{2}{10} = 1, 0.2\). Because \(p = \tan x\), so \( \tan x = 1, 0.2\).
03

Find the solution in the range

The period of the tangent function is \(\pi\) and it's defined for all real numbers. The solutions to the equations \(\tan x = 1\) and \(\tan x = 0.2\) will also be solutions to the equation \(\cot ^{2} x-6 \cot x+5=0\) and repeating every \(\pi\) radians.
04

Using the Inverse Tangent Function

In the next step use the inverse tangent function to find the angles of interest. For \(\tan x = 1\), \(x = \arctan(1)\) and\(x = \arctan(1) + k\pi\) where \(k\) is an integer. These are the angles \(x\) such that their tangent is \(1\). Also for \(\tan x = 0.2\), \(x = \arctan(0.2)\) and \(x = \arctan(0.2) + k\pi\) where \(k\) is an integer. These are the angles \(x\) such that their tangent is \(0.2\).
05

Find the specific solution in the range [0,2蟺).

Finally, list all values \(x = \arctan(1), \arctan(1) + \pi, \arctan(0.2), \arctan(0.2) + \pi \) that are in the interval [0,2蟺). The possible solutions are \(x = \frac{\pi}{4}, \frac{5\pi}{4}, \arctan(0.2), \arctan(0.2) + \pi \).

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Most popular questions from this chapter

Consider the function given by \(f(x)=3 \sin (0.6 x-2)\). (a) Approximate the zero of the function in the interval [0,6] (b) A quadratic approximation agreeing with \(f\) at \(x=5\) is \(g(x)=-0.45 x^{2}+5.52 x-13.70 .\) Use a graphing utility to graph \(f\) and \(g\) in the same viewing window. Describe the result. (c) Use the Quadratic Formula to find the zeros of \(g\). Compare the zero in the interval [0,6] with the result of part (a).

Use a graphing utility to approximate the solutions (to three decimal places) of the equation in the given interval. $$4 \cos ^{2} x-2 \sin x+1=0, \quad\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$$

Find the exact value of each expression. (a) \(\sin \left(135^{\circ}-30^{\circ}\right)\) (b) \(\sin 135^{\circ}-\cos 30^{\circ}\)

Find all solutions of the equation in the interval \([0,2 \pi)\). $$2 \cos ^{2} x+\cos x-1=0$$

Use inverse functions where needed to find all solutions of the equation in the interval \([0,2 \pi)\). $$\sec ^{2} x+\tan x-3=0$$
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