solve 3 csc theta/2 - 2 sqrt(3) = 0

[NEHA]

Solve the following equation given that 0 degrees < theta < 360 degrees:

. . .3csc theta/2 - 2sqrt(3) = 0

From which section is this? I lost the page in my book. And if you can help me out, I would be very grateful. Thank you!
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Edited by stapel -- Reason for edit: subject line, spelling, punctuation, etc

 

[stapel]

Since we aren't in your class, we of course have no way of know what section or page-number you should look at. But since you know this is a trig equation, and since you know that you've been instructed to solve it, you might want to consider looking in your Table of Contents or index for something along the lines of "solving trigonometric equations".

Once you have reviewed the topic and refamiliarized yourself with the basic terms and techniques, please attempt the exercise. If you get stuck, please reply showing what you have tried.

Thank you.

Eliz.

Edit: Ne'mind; the complete worked solution is provided in the next reply.

 

[soroban]

Hello, NEHA!

This problem may be in the section involving \(\displaystyle 30^o,\:60^o\), and \(\displaystyle 45^o\) angles.

Solve the following equation given that \(\displaystyle 0^o\,\leq\,\theta\,\leq\, 360^o\)

. . .\(\displaystyle 3\,\csc\left(\frac{\theta}{2}\right)\,-\,2\sqrt{3}\:=\: 0\)

We have: \(\displaystyle \L\,\csc\left(\frac{\theta}{2}\right)\:=\:\frac{2\sqrt{3}}{3} \:=\:\frac{2}{\sqrt{3}}\)

Then: \(\displaystyle \L\:\sin\left(\frac{\theta}{2}\right) \:=\:\frac{\sqrt{3}}{2}\)

Therefore: \(\displaystyle \L\:\frac{\theta}{2} \:=\:60^o,\:120^o\;\;\Rightarrow\;\;\fbox{\theta \:=\:120^o,\:240^o}\)