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Solve the equation from $0\leq x < 360$

$\sqrt{3}\tan(3x)=1,$

I have tried to solve this equation by dividing the square root into both sides, and then making $3x$ equal to inverse $\tan$ of 1/$\sqrt{3}$. I only get $3x=30$ and $x=10$. I do not believe that this is right.

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    $\begingroup$ It is correct, but it is not the only solution (provided that you mean $10^\circ$). $\endgroup$
    – Andrew Chin
    Commented Jan 17, 2021 at 1:13

2 Answers 2

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Just divide by $\sqrt{3}$ to get tan(3x) = 1/$\sqrt{3}$ then do $3x=tan^{-1}(1/\sqrt{3})+180^{o}k$ for k=0,1,2,3... So x = 10, 70, 130,..

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  • $\begingroup$ You can do x^{\circ} ($x^{\circ}$) for a degree symbol. $\endgroup$
    – Joshua Wang
    Commented Jan 17, 2021 at 1:18
  • $\begingroup$ Also, don't forget negative $k$. $\endgroup$
    – Joshua Wang
    Commented Jan 17, 2021 at 1:21
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Let $y = 3x$, so $0^{\circ}\leq y < 1080^{\circ}$. Then, we must have:

$$\sqrt{3}\tan(y) = 1$$

$$\tan(y) = \frac{1}{\sqrt{3}}$$

Because $\tan(y) = \frac{1}{\sqrt{3}}$ whenever $y = 30^{\circ} + n\cdot 180^{\circ}$ for any integer $n$, we find that the solutions satisfying $0^{\circ}\leq y < 1080^{\circ}$ are $y= 30^{\circ},\ 210^{\circ},\ 390^{\circ},\ 570^{\circ},\ 750^{\circ},\ 930^{\circ}$. Then, we have:

$$x = \frac{y}{3} = \boxed{10^{\circ},\ 70^{\circ},\ 130^{\circ},\ 190^{\circ},\ 250^{\circ},\ 310^{\circ}}$$

Note that the solutions all differ by $60^{\circ}$, one third the period of $\tan$.

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