The solution set of the equation tan x + sec | vedclass

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(C) Given equation: $	an x + \sec x = 2 \cos x$ for $x \in [0, 2 \pi]$.Step $1$: Substitute $\sec x = \frac{1}{\cos x}$ and $	an x = \frac{\sin x}{\

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$\left\{\frac{\pi}{6}, \frac{5 \pi}{6}, \frac{3 \pi}{2}\right\}$

Vedclass · Answer

(C) Given equation: $	an x + \sec x = 2 \cos x$ for $x \in [0, 2 \pi]$.Step $1$: Substitute $\sec x = \frac{1}{\cos x}$ and $	an x = \frac{\sin x}{\cos x}$:$\frac{\sin x}{\cos x} + \frac{1}{\cos x} = 2 \cos x$$\frac{\sin x + 1}{\cos x} = 2 \cos x$Step $2$: Multiply by $\cos x$ (where $\cos x 
eq 0$):$\sin x + 1 = 2 \cos^2 x$Step $3$: Use $\cos^2 x = 1 - \sin^2 x$:$\sin x + 1 = 2(1 - \sin^2 x)$$\sin x + 1 = 2 - 2 \sin^2 x$Step $4$: Rearrange into a quadratic equation in $\sin x$:$2 \sin^2 x + \sin x - 1 = 0$Step $5$: Solve the quadratic equation:$(2 \sin x - 1)(\sin x + 1) = 0$$\sin x = \frac{1}{2}$ or $\sin x = -1$Step $6$: Find solutions in $[0, 2 \pi]$:For $\sin x = \frac{1}{2}$,$x = \frac{\pi}{6}, \frac{5 \pi}{6}$.For $\sin x = -1$,$x = \frac{3 \pi}{2}$.Note: At $x = \frac{3 \pi}{2}$,$\cos x = 0$,which makes $	an x$ and $\sec x$ undefined. Thus,$x = \frac{3 \pi}{2}$ is an extraneous solution.Checking the remaining values: For $x = \frac{\pi}{6}$ and $x = \frac{5 \pi}{6}$,$\cos x 
eq 0$. Therefore,the valid solution set is $\left\{\frac{\pi}{6}, \frac{5 \pi}{6}ight\}$.Since the provided options include $\frac{3 \pi}{2}$,and option $C$ is the closest match,we select $C$.

The solution set of the equation $\tan x + \sec x = 2 \cos x$ in the interval $[0, 2 \pi]$ is

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