# Trigonometric equation $\tan4x=\sqrt 3,\qquad 0\leq x \lt \pi$

$$\tan4x=\sqrt 3,\qquad 0\leq x \lt \pi$$

4 solutions: $$\frac{\pi}{12}, \frac{\pi}{3}, \frac{7\pi}{12}, \frac{5\pi}{6}$$

or 3 solutions: $$\frac{\pi}{12}, \frac{\pi}{3}, \frac{7\pi}{12}$$ (text suggested this and no clues about why $$\frac{5\pi}{6}$$ was excluded

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It was excluded because books and "official" solutions sometimes have mistakes. And they should have given all the answers with denominator $12$, simplifying hides structure. – André Nicolas Sep 28 '12 at 17:51
verifying textbook now, since it is quite credible, i wonder if i made that wrong – Paul Smith Sep 28 '12 at 18:00

Ah, yes, the original formulation of the question shows why the book only has three answers. If you plug in $\frac{5\pi}{6}$ then it gets multiplied by the $3$ in two of the tangent terms and $$\tan({3(\frac{5\pi}{6})})=\tan{\frac{5\pi}{2}}$$ is not defined.

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ya, thanks, i got that now – Paul Smith Sep 29 '12 at 12:15

Edit: forget what I said, it should be 4 solutions, as the comment says, ($20\pi/6$) is ($< 4\pi$)

I put it in my mind as $25\pi/6$

$\text{let}\; \theta = 4x$

$0 < \theta < 4\pi$

If you now solve for $\theta$, you will get four solutions.

But $20/6 < 4$. – SiliconCelery Sep 28 '12 at 17:49
I am not following, $$\frac {20\pi}{6} = 3.3333333\pi \lt 4\pi$$? – Paul Smith Sep 28 '12 at 17:51
i count 4 if I solve for $$\theta$$? $$4\pi$$ is two revolution, the $$1^{st}$$ quadrant, $$3^{rd}$$ quadrant, $$5^{th}$$ quadrant, $$7^{th}$$ quardrant – Paul Smith Sep 28 '12 at 17:54
i don't know if i made compound angles right, the original question is $$tanx+tan3x+\sqrt{3}tanxtan3x=tanxtan3x$$, same range for $$\theta$$, does it justifty 3 answers? – Paul Smith Sep 28 '12 at 17:57