Solve tan(x) + sec(x) = sqrt{3} for x in [0, | vedclass

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(B) Given the equation: $	an(x) + \sec(x) = \sqrt{3}$.We know the identity $\sec^2(x) - 	an^2(x) = 1$,which implies $(\sec(x) - 	an(x))(\sec(x) + \

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$\frac{\pi}{6}$

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(B) Given the equation: $	an(x) + \sec(x) = \sqrt{3}$.We know the identity $\sec^2(x) - 	an^2(x) = 1$,which implies $(\sec(x) - 	an(x))(\sec(x) + 	an(x)) = 1$.Substituting the given value: $(\sec(x) - 	an(x))(\sqrt{3}) = 1$,so $\sec(x) - 	an(x) = \frac{1}{\sqrt{3}}$.Adding the two equations:$(	an(x) + \sec(x)) + (\sec(x) - 	an(x)) = \sqrt{3} + \frac{1}{\sqrt{3}}$$2\sec(x) = \frac{3+1}{\sqrt{3}} = \frac{4}{\sqrt{3}}$ $\Rightarrow \sec(x) = \frac{2}{\sqrt{3}}$ $\Rightarrow \cos(x) = \frac{\sqrt{3}}{2}$.For $x \in [0, 2\pi]$,the solutions are $x = \frac{\pi}{6}$ and $x = \frac{11\pi}{6}$.Checking the original equation: $	an(\frac{\pi}{6}) + \sec(\frac{\pi}{6}) = \frac{1}{\sqrt{3}} + \frac{2}{\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3}$.Thus,$x = \frac{\pi}{6}$ is a valid solution.

Solve $\tan(x) + \sec(x) = \sqrt{3}$ for $x \in [0, 2\pi]$.

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