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: 44


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

Work Step by Step

$$2\cos^2\theta+2\cos\theta+1=0$$ over interval $[0^\circ,360^\circ)$ 1) Solve the equation: $$2\cos^2\theta+2\cos\theta+1=0$$ Here we need to consider the equation a quadratic formula. In fact, if we take $\cos\theta$ as $x$, we would get $$2x^2+2x+1=0$$ with $a=2, b=2$ and $c=1$ - Calculate $\Delta$: $\Delta=b^2-4ac=2^2-4\times2\times1=4-8=-4\lt0$ $\Delta\lt0$ means that there would be no values of $x\in R$ which can satisfy the given equation. That also means there would be no values of $\theta\in[0^\circ,360^\circ)$ which can satisfy the original equation $$2\cos^2\theta+2\cos\theta+1=0$$ Therefore, the solution set to this problem is $$\{\varnothing\}$$
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