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Chapter 5: Problem 72

Use inverse functions where needed to find all solutions of the equation in the interval \([0,2 \pi)\). $$\sec ^{2} x+2 \sec x-8=0$$

Short Answer

Expert verified
There are no solutions to this equation in the given range from 0 to \(2 \pi\).

Step by step solution

01

Identify the Form of the Given Equation

The given equation \(\sec ^{2} x+2 \sec x-8=0\) looks like a quadratic equation in term of sec(x).
02

Change the form of the equation

By setting \(y = \sec x\), we transform the equation into the form \(y^2 + 2y - 8=0\). Now we can solve for y using the quadratic formula.
03

Apply the Quadratic Formula

The quadratic formula is \((-b\pm \sqrt{b^{2}-4ac})/2a\). In this case, a=1, b=2, and c=-8. Substituting these values in, we get \(y=(-2\pm \sqrt{2^2-4(1)(-8)})/2(1)= -2\pm \sqrt{36}/2=-2\pm 6 \). Therefore, our solutions are \(y=-8\) and \(y=4\).
04

Go Back to the Original Variable

Remembering y = sec(x), our solutions, in terms of x, are when sec(x)=-8 and sec(x)=4. These represent x = sec^{-1}(-8) and x = sec^{-1}(4). We need to find these values in the range from 0 to \(2 \pi\) but since the secant function \(-1 鈮� sec(x) 鈮� -1\) or \(1 鈮� sec(x) 鈮� 鈭瀄), there are no solutions in the given range.

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Most popular questions from this chapter

(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 $$f(x)=\sin x \cos x$$ Trigonometric Equation $$-\sin ^{2} x+\cos ^{2} x=0$$

Find the \(x\) -intercepts of the graph. $$y=\sec ^{4}\left(\frac{\pi x}{8}\right)-4$$

a sharpshooter intends to hit a target at a distance of 1000 yards with a gun that has a muzzle velocity of 1200 feet per second (see figure). Neglecting air resistance, determine the gun's minimum angle of elevation \(\theta\) if the range \(r\) is given by $$r=\frac{1}{32} v_{0}^{2} \sin 2 \theta$$

Find all solutions of the equation in the interval \([0,2 \pi)\). $$\csc x+\cot x=1$$

Use the Quadratic Formula to solve the equation in the interval \([0,2 \pi)\). Then use a graphing utility to approximate the angle \(x\). $$\tan ^{2} x+3 \tan x+1=0$$
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