Chapter 5 - Section 5.5 - Trigonometric Equations - Exercise Set - Page 703: 19

The solutions of the equation are $x=\frac{5\pi }{6}+2n\pi$ or $x=\frac{7\pi }{6}+2n\pi$, where n is any integer.

Work Step by Step

\begin{align} & 2\cos x+\sqrt{3}=0 \\ & 2\cos x=0-\sqrt{3} \\ & \cos x=-\frac{\sqrt{3}}{2} \end{align} $\cos \frac{\pi }{6}=\frac{\sqrt{3}}{2}$, and the solutions of $\cos x=-\frac{\sqrt{3}}{2}$ in [0,2 $\pi$ ) are: \begin{align} & x=\pi -\frac{\pi }{6} \\ & =\frac{6\pi }{6}-\frac{\pi }{6} \\ & =\frac{5\pi }{6} \end{align} \begin{align} & x=\pi +\frac{\pi }{6} \\ & =\frac{6\pi }{6}+\frac{\pi }{6} \\ & =\frac{7\pi }{6} \end{align} The period of the sine function is $2\pi$, and the solutions are given by: $x=\frac{5\pi }{6}+2n\pi$ or $x=\frac{7\pi }{6}+2n\pi$, where n is any integer.

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