## Trigonometry (11th Edition) Clone

Published by Pearson

# Chapter 6 - Inverse Circular Functions and Trigonometric Equations - Section 6.2 Trigonometric Equations I - 6.2 Exercises - Page 273: 42

#### Answer

The solution set to this problem is $$\theta\in\{38.4^\circ, 104.8^\circ,218.4^\circ,284.8^\circ\}$$

#### Work Step by Step

$$3\cot^2\theta-3\cot\theta-1=0$$ over interval $[0^\circ,360^\circ)$ 1) Solve the equation: $$3\cot^2\theta-3\cot\theta-1=0$$ Here we need to consider the equation a quadratic formula. In fact, if we take $\cot\theta$ as $x$, we would get $$3x^2-3x-1=0$$ with $a=3, b=-3$ and $c=-1$ - Calculate $\Delta$: $\Delta=b^2-4ac=(-3)^2-4\times3\times(-1)=9-(-12)=21$ - Calculate $x$, or in other words, $\cot\theta$: $$x=\cot\theta=\frac{-b\pm\sqrt\Delta}{2a}=\frac{3\pm\sqrt{21}}{6}$$ 2) Apply the inverse function: - For $\cot\theta=\frac{3+\sqrt{21}}{6}$: $$\cot\theta=\frac{3+\sqrt{21}}{6}\approx1.264$$ which means $$\theta=\cot^{-1}(1.264)$$ $$\theta\approx38.4^\circ\hspace{1cm}\text{or}\hspace{1cm}\theta\approx218.4^\circ$$ - For $\cot\theta=\frac{3-\sqrt{21}}{6}$: $$\cot\theta=\frac{3-\sqrt{21}}{6}\approx-0.264$$ which means $$\theta=\cot^{-1}(-0.264)$$ $$\theta\approx104.8^\circ\hspace{1cm}\text{or}\hspace{1cm}\theta\approx284.8^\circ$$ Therefore, $$\theta\in\{38.4^\circ, 104.8^\circ,218.4^\circ,284.8^\circ\}$$ In other words, the solution set to this problem is $$\theta\in\{38.4^\circ, 104.8^\circ,218.4^\circ,284.8^\circ\}$$

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