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    Find the general solution of equation tanx=13tanx=\frac{1}{\sqrt3}tanx=3​1​.
    Question

    Find the general solution of equation tanx=13tanx=\frac{1}{\sqrt3}.

    A.

    x=nπ+5π6x=nπ+\frac{5π}{6}​​

    B.

    x=nπ+π3x=nπ+\frac{π}{3}​​

    C.

    x=nπ+π6x=nπ+\frac{π}{6}​​

    D.

    x=nπ+2π3x=nπ+\frac{2π}{3}​​

    Correct option is C

    First, find the principal solution (i.e., the smallest positive angle x that satisfies the equation).We know that:tan(π6)=13Thus, one solution is:x=π6Determine the General SolutionThe tangent function has a period of π, meaning:tanx=tan(x+nπ)for any integer n.Therefore, the general solution is:x=π6+nπwhere nZ (i.e., n=0,±1,±2,)\begin{aligned}&\text{First, find the principal solution (i.e., the smallest positive angle } x \text{ that satisfies the equation).} \\&\text{We know that:} \\&\tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}} \\&\text{Thus, one solution is:} \\&x = \frac{\pi}{6} \\\\&\textbf{Determine the General Solution} \\&\text{The tangent function has a period of } \pi, \text{ meaning:} \\&\tan x = \tan(x + n\pi) \quad \text{for any integer } n. \\&\text{Therefore, the general solution is:} \\&x = \frac{\pi}{6} + n\pi \quad \text{where } n \in \mathbb{Z} \text{ (i.e., } n = 0, \pm1, \pm2, \ldots \text{)}\end{aligned}​​

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