Question 808848
<font face="Times New Roman" size="+2">


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \tan\varphi\ =\ \frac{\sin\varphi}{\cos\varphi}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \csc\varphi\ =\ \frac{1}{\sin\varphi}]


So


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \tan\theta\csc\theta\ -\ \sqrt{3}\csc\theta\ =\ 0]


is equivalent to


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1}{\cos\theta}\ -\ \frac{\sqrt{3}}{\sin\theta}\ =\ 0]


Add


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\sqrt{3}}{\sin\theta}]


to both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1}{\cos\theta}\ =\ \frac{\sqrt{3}}{\sin\theta}]


Cross multiply:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin\theta\ =\ \sqrt{3}\cos\theta]


Square both sides


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin^2\theta\ =\ 3\cos^2\theta]


Use the Pythagorean identity


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin^2\theta\ =\ 3\ -\ 3\sin^2\theta]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4\sin^2\theta\ =\ 3]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin\theta\ =\ \pm\frac{\sqrt{3}}{2}]


From which we get, using the unit circle, that


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \theta\ =\ \frac{\pi}{3},\,\frac{2\pi}{3},\,\frac{4\pi}{3},\,\frac{5\pi}{3}]


Which then implies that *[tex \Large \cos\theta\ =\ \pm\frac{1}{2}]


But if sine and cosine are opposite signs, the original equation fails.  Hence, the solution set is confined to Quadrants I and III where the signs of sine and cosine are the same.  And the solution set is therefore:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left{\theta\ \in\ \mathbb{R}\ |\ \theta =\ \frac{\pi}{3} \ \large\vee\LARGE\ \theta\ =\ \frac{4\pi}{3}\right}]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
<div style="text-align:center"><a href="http://outcampaign.org/" target="_blank"><img src="http://cdn.cloudfiles.mosso.com/c116811/scarlet_A.png" border="0" alt="The Out Campaign: Scarlet Letter of Atheism" width="143" height="122" /></a></div>
</font>