## 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: 57

#### Answer

The solution set is $$\{0.8751+2\pi n, 2.2665+2\pi n,3.5908+2\pi n, 5.834+2\pi n, n\in Z\}$$

#### Work Step by Step

$$\sin x(3\sin x-1)=1$$ 1) Solve the equation over the interval $[0,2\pi)$ $$\sin x(3\sin x-1)=1$$ $$3\sin^2x-\sin x-1=0$$ Consider the equation as a quadratic formula, with $a=3, b=-1, c=-1$ - Calculate $\Delta$: $\Delta=b^2-4ac=(-1)^2-4\times3\times(-1)=1+12=13$ - Find out $\sin x$: $$\sin x=\frac{-b\pm\sqrt\Delta}{2a}=\frac{1\pm\sqrt{13}}{6}$$ - For $\sin x=\frac{1+\sqrt{13}}{6}$. $x$ would be $$x=\sin^{-1}\frac{1+\sqrt{13}}{6}$$ $$x\approx0.8751$$ (Be careful with degrees and radians. Here we need to use radians.) However, over the interval $[0,2\pi)$, there is one more value of $x$ where $\sin x=\frac{1+\sqrt{13}}{6}$, which is $x=\pi-0.8751\approx2.2665$ In total, $x\in\{0.8751, 2.2665\}$ - For $\sin x=\frac{1-\sqrt{13}}{6}$. $x$ would be $$x=\sin^{-1}\frac{1-\sqrt{13}}{6}$$ $$x\approx-0.4492$$ In the interval $[0,2\pi)$, that equals to $x\approx-0.4492+2\pi\approx5.834$ However, over the interval $[0,2\pi)$, there is one more value of $x$ where $\sin x=\frac{1-\sqrt{13}}{6}$, which is $x=\pi-(-0.4492)\approx3.5908$ In total, $x\in\{3.5908, 5.834\}$ Therefore, overall, $$x=\{0.8751, 2.2665,3.5908, 5.834\}$$ 2) Solve the equation for all solutions - The integer multiples of the period of the sine function is $2\pi$. - We apply it to each solution found in step 1. The results are the solution set. So the solution set is $$\{0.8751+2\pi n, 2.2665+2\pi n,3.5908+2\pi n, 5.834+2\pi n, n\in Z\}$$

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