How would one go about finding the exact value of $\theta$ in the following:$\sqrt{3}\tan \theta -1 =0$? I am unsure of how to begin this question. Any help would be appreciated!
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$\begingroup$ are you familiar with $\arctan$ function? $\endgroup$– AlexJun 18, 2013 at 16:15
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$\begingroup$ I am not familiar with that function $\endgroup$– ComradeYakovJun 18, 2013 at 16:16
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$\begingroup$ @Tim I meant the first, sorry for the confusion. It isn't cubed $\endgroup$– ComradeYakovJun 18, 2013 at 16:17
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2 Answers
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If $\sqrt{3}\tan \theta - 1= 0$ then $\sqrt{3}\tan\theta = 1$ and $\tan\theta = 1/\sqrt{3}$. Hence $\theta =\arctan(1/\sqrt{3}) = \frac{\pi}{6}$.
Notice that the graph $y = \tan\theta$ repeats every $\pi$-radians. Hence, all solutions are given by
$$\theta = \frac{\pi}{6} + \pi n$$
where $n$ is any whole number.
answered Jun 18, 2013 at 16:18
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Hint: $\sqrt{3}\tan \theta -1 =0\implies \sqrt{3}\tan \theta =1\implies \tan \theta=1/\sqrt{3}$
Recall from SOH CAH TOA, that $\tan \theta=\text{opposite}/\text{adjacent}$, so in this case $1$ is the opposite side and $\sqrt{3}$ is the adjacent side.
The above is a special triangle. Since you know that $\tan \theta=1/\sqrt{3}$, it follows that the angle $\theta=\pi/6$. However since $\tan \theta$ is a trigonometric periodic function, that value will come up again, and again. (Just like $\sin0=\sin2\pi).$
$\tan \theta$ has a period of $\pi.$
So to be fully accurate, $\theta=\pi/6 +n\pi, n \in \mathbb{Z}$
answered Jun 18, 2013 at 16:15
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$\begingroup$ Thank you! So theta= pi/4(+-)pi(n), 5pi/4(+-) is not a solution? $\endgroup$ Jun 18, 2013 at 16:31
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$\begingroup$ theta= pi/4(+-)pi(n), 5pi/4(+-) is not a solution. $\endgroup$ Jun 18, 2013 at 16:35
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1$\begingroup$ $2\cos^{2} \theta+\cos \theta-1=0 \implies 2\cos^{2}+\cos \theta =1 \implies \cos \theta (2\cos \theta+1)=1\implies \cos \theta =1 \implies \theta =2n\pi, n\in \mathbb{Z}$ However, $\cos \theta(2\cos \theta +1)=1$ also implies that $2\cos \theta+1=1 \implies 2\cos \theta =0 \implies \cos \theta =0 \implies \theta = (2n-1)/\pi, n \in \mathbb{Z}$ $\endgroup$ Jun 18, 2013 at 17:07