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Chapter 5: Problem 64

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

Short Answer

Expert verified
The solutions within the interval \([0,2\pi)\) to the equation \(\tan^2x - \tan x - 2 = 0\) are \(x = \arctan(2)\), \(x = \pi + \arctan(2)\) and \(x = \pi + \arctan(-1)\).

Step by step solution

01

Write the Equation in Quadratic Form

Consider \(\tan x\) as \(y\). The equation can be rewritten as \(y^2 - y - 2 = 0\).
02

Solve the Quadratic Equation

Factorize the quadratic equation to find the roots. Factors of \(y^2 - y - 2 = 0\) are \((y - 2)(y + 1) = 0\). Setting each factor equal to zero gives \(y = 2\) and \(y = -1\).
03

Find the Corresponding Angles for Tangent Values

Substitute \(y\) back with \(\tan x\), to give two values: \(\tan x = 2\) and \(\tan x = -1\). The angles corresponding to these tangent values should be determined using the inverse tangent function, noting that we need to cover all possible angles in the interval \([0,2\pi)\).
04

Apply Inverse Tangent Function

Using inverse tangent function, for \(\tan x = 2\), we get \(x = \arctan(2)\). However, the tangent function has a period of \(\pi\), so an additional solution in the interval \([0,2\pi)\) is \(x = \pi + \arctan(2)\). Similarly, for \(\tan x = -1\), we have \(x = \arctan(-1)\) and \(x = \pi + \arctan(-1)\). However \(\arctan(-1)\) is negative and doesn't fall in the interval \([0,2\pi)\) but if we add \(\pi\) to \(\arctan(-1)\), it will fall in the interval \([0,2\pi)\). Therefore discard \(\arctan(-1)\) and take \(x = \pi + \arctan(-1)\).

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Most popular questions from this chapter

Solve the multiple-angle equation. $$\sin 2 x=-\frac{\sqrt{3}}{2}$$

Use a graphing utility to approximate the solutions (to three decimal places) of the equation in the interval \([0,2 \pi)\). $$x \cos x-1=0$$

Determine whether the statement is true or false. Justify your answer. If you correctly solve a trigonometric equation to the statement \(\sin x=3.4\), then you can finish solving the equation by using an inverse function.

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

(a) use a graphing utility to graph the function and approximate the maximum and minimum points on the graph in the interval \([0,2 \pi),\) and (b) solve the trigonometric equation and demonstrate that its solutions are the \(x\) -coordinates of the maximum and minimum points of \(f .\) (Calculus is required to find the trigonometric equation.) Function $$f(x)=\cos ^{2} x-\sin x$$ Trigonometric Equation $$-2 \sin x \cos x-\cos x=0$$
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