# Solve for θ over the given domain

• Jul 16th 2008, 06:09 AM
sweetG
Solve for θ over the given domain
√3 sin 2θ = cos 2θ
Domain: 0o ≤ θ ≤ 360o
• Jul 16th 2008, 08:16 AM
Soroban
Hello, sweetG!

Quote:

$\displaystyle \sqrt{3}\sin2\theta \:=\:\cos2\theta\qquad 0^o \leq \theta \leq 360^o$
Divide by $\displaystyle \sqrt{3}\cos2\theta\!:\;\;\frac{\sin2\theta}{\cos2 \theta} \:=\:\frac{1}{\sqrt{3}} \quad\Rightarrow\quad \tan2\theta \:=\:\frac{1}{\sqrt{3}}$

Then: .$\displaystyle 2\theta\;=\;30^o,\: 210^o,\: 390^o,\: 570^o$

Therefore: .$\displaystyle \theta \;=\;15^o,\:105^o,\:195^o,\:285^o$

• Jul 18th 2008, 07:36 AM
sweetG
Thanks so much Soroban that really helped a lot :)
But I've forgotten how to do one thing......how did you get the answers that come after 30 degrees i.e. 210, 390 and 570 degrees. I've forgotten how do these types of questions:(
• Jul 18th 2008, 09:11 AM
Simplicity
Quote:

Originally Posted by sweetG
Thanks so much Soroban that really helped a lot :)
But I've forgotten how to do one thing......how did you get the answers that come after 30 degrees i.e. 210, 390 and 570 degrees. I've forgotten how do these types of questions:(

The $\displaystyle \tan \theta$ function has a period of $\displaystyle 180^{\circ}$ hence when you add $\displaystyle 180^{\circ}$, you get the value for $\displaystyle \theta$ which will satisfy your equation. There are infinite solution hence questions usually provide a range (in this case $\displaystyle 0^o \leq \theta \leq 360^o$).
• Jul 19th 2008, 07:07 AM
sweetG
Thanks Soroban and Air that really helped and now I'm a lot more clearer on how to do these types of questions :)