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The principal solutions of $\sqrt{3} \sec x+2=0$ are
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2767 Upvotes
Verified Answer
The correct answer is:
$\frac{5 \pi}{6}, \frac{7 \pi}{6}$
$$
\begin{aligned}
& \sqrt{3} \sec x+2=0 \\
& \therefore \sec x=\frac{-2}{\sqrt{3}} \Rightarrow \cos x=\frac{-\sqrt{3}}{2} \\
& \therefore \cos x=\cos \left(\pi-\frac{\pi}{6}\right)=\cos \left(\pi+\frac{\pi}{6}\right) \Rightarrow x=\frac{5 \pi}{6}, \frac{7 \pi}{6}
\end{aligned}
$$
\begin{aligned}
& \sqrt{3} \sec x+2=0 \\
& \therefore \sec x=\frac{-2}{\sqrt{3}} \Rightarrow \cos x=\frac{-\sqrt{3}}{2} \\
& \therefore \cos x=\cos \left(\pi-\frac{\pi}{6}\right)=\cos \left(\pi+\frac{\pi}{6}\right) \Rightarrow x=\frac{5 \pi}{6}, \frac{7 \pi}{6}
\end{aligned}
$$
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