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

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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    verifying textbook now, since it is quite credible, i wonder if i made that wrong2012-09-28

2 Answers 2

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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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    y$a$, thanks, i got that now2012-09-29
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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.

Sorry about that....

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    The one you wrote in the comment (the original question in the textbook) and the one in the main question. Maybe you could post the steps you took to go from the original question to $\tan 4x = \sqrt{3}$2012-09-29