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Chapter 5: Problem 66

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

Short Answer

Expert verified
So, we have three solutions in the interval \([0, 2\pi)\), which are \(x = \frac{\pi}{4}, \frac{5\pi}{4}, \frac{5\pi}{4}\).

Step by step solution

01

Rewrite the Equation

Using the Pythagorean identity, replace \(\sec ^{2} x\) with \(1 + \tan ^{2} x\), rewriting the equation as \(1 + \tan^2x + \tan x - 3 = 0.\) This simplifies to \(\tan^2x + \tan x - 2 = 0\).
02

Solve the Quadratic Equation

We can think of \(\tan x\) as the variable and treat the equation as a standard quadratic equation. Following the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}} {2a}\), we get roots as \(\tan x = -2, 1\).
03

Find the Corresponding Angles

To solve for \(x\), we need to find the angles that yield these tangent values in the interval \([0,2\pi)\). For \(\tan x = -2\), the solutions are \(\frac{5\pi}{4}, \frac{9\pi}{4}\). However, since \(2\pi < \frac{9\pi}{4} < \frac{10\pi}{4}\), we discard the value \(\frac{9\pi}{4}\). For \(\tan x = 1\), the solutions are \(\frac{\pi}{4}, \frac{5\pi}{4}\).
04

Check and Verify the solutions

Finally, all potential solutions should be tested in the original equation \(\sec ^{2} x+\tan x-3 = 0\) to ensure they are valid in the given interval \([0,2\pi)\).

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