MHB Can You Solve the Existence of $x$ in this Trigonometric Equation?

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    2017
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The discussion centers around proving the existence of a natural number \( x \) within the range of 1 to 89 that satisfies the equation \( \sqrt{3}\tan x^\circ - 1 = \sec 20^\circ \). Participants are encouraged to engage with the Problem of the Week (POTW) and follow the provided guidelines for submissions. Several members, including Opalg, kaliprasad, and greg1313, successfully provided correct solutions to the problem. The thread highlights the collaborative nature of solving mathematical challenges within the community. Engaging with such problems fosters deeper understanding and application of trigonometric concepts.
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Here is this week's POTW:

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Prove that there exists $x\in \Bbb{N}$, where $1\le x \le 89$ such that $\sqrt{3}\tan x^\circ-1=\sec 20^\circ$.

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!
 
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Congratulations to the following members for their correct solution: (Smile)

1. Opalg
2. kaliprasad
3. greg1313

Solution from Opalg:
A quick check on a calculator suggests that the answer should be $x=50$. With that in mind, use the addition formulas for sin and cos, and the values of $\sin 30^\circ$ and $\cos 30^\circ$, to get $$\tan 50^\circ = \frac{\sin(30^\circ+20^\circ)} {\cos(30^\circ+20^\circ)} = \frac{\frac12(\cos 20^\circ + \sqrt3\sin 20^\circ)}{\frac12(\sqrt3\cos 20^\circ - \sin 20^\circ)},$$ $$ \sqrt3\tan 50^\circ - 1 = \frac{\sqrt3(\cos 20^\circ + \sqrt3\sin 20^\circ) - (\sqrt3\cos 20^\circ - \sin 20^\circ)}{\sqrt3\cos 20^\circ - \sin 20^\circ} = \frac{4\sin 20^\circ }{\sqrt3\cos 20^\circ - \sin 20^\circ}.\qquad(1)$$ The denominator in (1) is $$\sqrt3\cos 20^\circ - \sin 20^\circ = 2\cos 50^\circ = 2\sin40^\circ = 4\sin20^\circ\cos20^\circ.$$ It follows from (1) that $ \sqrt3\tan 50^\circ - 1 = \dfrac{4\sin 20^\circ }{4\sin20^\circ\cos20^\circ} = \dfrac1{\cos20^\circ} = \sec20^\circ$ , as required.
 
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