# Trig Identity Math Problem $\sqrt{3} \tan(3x)=1$ from $0\leq x<360$

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.

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

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,..

• You can do x^{\circ} ($x^{\circ}$) for a degree symbol. Commented Jan 17, 2021 at 1:18
• Also, don't forget negative $k$. Commented Jan 17, 2021 at 1:21

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 $$\tan$$.