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(A) Given $\sec x = 2$. Since $\sec x = \frac{1}{\cos x}$,we have $\cos x = \frac{1}{2}$. We know that $\cos \frac{\pi}{3} = \frac{1}{2}$. Since $\cos

Vedclass · Accepted Answer

$x = \frac{\pi}{3}, \frac{5\pi}{3}$ and $x = 2n\pi \pm \frac{\pi}{3}, n \in \mathbb{Z}$

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(A) Given $\sec x = 2$. Since $\sec x = \frac{1}{\cos x}$,we have $\cos x = \frac{1}{2}$. We know that $\cos \frac{\pi}{3} = \frac{1}{2}$. Since $\cos x$ is positive in the first and fourth quadrants,the principal solutions are $x = \frac{\pi}{3}$ and $x = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}$. For the general solution,if $\cos x = \cos \alpha$,then $x = 2n\pi \pm \alpha$,where $n \in \mathbb{Z}$. Here $\alpha = \frac{\pi}{3}$,so the general solution is $x = 2n\pi \pm \frac{\pi}{3}$,where $n \in \mathbb{Z}$.

Find the principal and general solutions of the equation $\sec x = 2$.

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