Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 6 - Inverse Circular Functions and Trigonometric Equations - Section 6.2 Trigonometric Equations I - 6.2 Exercises - Page 273: 23


The solution set to this problem is $$\{\pi\}$$

Work Step by Step

$$\cos^2x+2\cos x+1=0$$ over interval $[0,2\pi)$ 1) Solve the equation: $$\cos^2x+2\cos x+1=0$$ $$(\cos x+1)^2=0$$ $$\cos x+1=0$$ $$\cos x=-1$$ 2) Apply the inverse function: $\cos x=-1$: $\cos x\lt0$ means that the angle of $x$ lies either in quadrant II or III. In fact, looking at the unit circle, we can easily figure out at only $x=\pi$, $\cos x=-1$ Therefore, $$x=\pi$$ In other words, the solution set to this problem is $$\{\pi\}$$
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