find all solutions of the equation in the interval $0,2\\pi)$. \n\n$\\sqrt{3}\\sec\\theta + 2 = 0$ \n\nwrite…

Explanation:

Step1: Solve for $\sec\theta$

Given $\sqrt{3}\sec\theta + 2 = 0$, we first isolate $\sec\theta$. Subtract 2 from both sides: $\sqrt{3}\sec\theta=-2$. Then divide both sides by $\sqrt{3}$: $\sec\theta =-\frac{2}{\sqrt{3}}$. Since $\sec\theta=\frac{1}{\cos\theta}$, we have $\cos\theta =-\frac{\sqrt{3}}{2}$.

Step2: Find $\theta$ in $[0,2\pi)$

We know that $\cos\theta =-\frac{\sqrt{3}}{2}$. The general solution for $\cos\theta = a$ is $\theta = 2k\pi\pm\cos^{-1}(a)$. For $a =-\frac{\sqrt{3}}{2}$ and $k = 0$, $\cos^{-1}\left(\frac{\sqrt{3}}{2}\right)=\frac{\pi}{6}$. Using the formula $\theta=\pi\pm\frac{\pi}{6}$. When $\theta=\pi+\frac{\pi}{6}=\frac{7\pi}{6}$ and when $\theta=\pi - \frac{\pi}{6}=\frac{5\pi}{6}$.

Answer:

$\frac{5\pi}{6},\frac{7\pi}{6}$