Problem 79

Question

In Exercises \(79-84,\) (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.) $$\begin{array}{ll} \qquad {\text { Function }} & {\text { Trigonometric Equation }} \\ {f(x)=\sin ^{2} x+\cos x} & {2 \sin x \cos x-\sin x=0}\end{array}$$

Step-by-Step Solution

Verified
Answer
The function \( f(x) = \sin^2 x + \cos x \) has maxima at \( x = \frac{\pi}{3}, \frac{5\pi}{3} \) and minima at \( x = 0, \pi \) in the interval [0,2). Correspondingly, the solutions to the equation \( 2 \sin x \cos x - \sin x = 0 \) are \( x = 0, \pi, \frac{\pi}{3}, \frac{5\pi}{3} \). This confirms that the solutions to the equation match the x-coordinates of the maximum and minimum points of the function.
1Step 1: Graphing the function
Plot the function \( f(x) = \sin^2 x + \cos x \) using a graphing utility. Now, identify the maximum and minimum points in the interval [0,2). The maximum and minimum points in the graph correspond to the maximum and minimum values of the function in the interval.
2Step 2: Solving the trigonometric equation
Now solve the given trigonometric equation \( 2 \sin x \cos x - \sin x = 0 \). Factor out \(\sin x\) to obtain \(\sin x (2\cos x - 1) = 0\). Setting each factor equal to zero we get \(\sin x = 0\) and \( 2\cos x - 1 = 0\). The solutions to these equations are \( x = 0, \pi, \frac{\pi}{3}, \frac{5\pi}{3} \) in our interval.
3Step 3: Connecting the solutions to the graph
The solutions of the trigonometric equation should correspond to the x-coordinates of the maximum and minimum points on the graph. The maximum points occur when \( x = \frac{\pi}{3}, \frac{5\pi}{3} \) and the minimum points occur when \( x = 0, \pi \). This should match the observations made from the graph in Step 1.