Thread: Solve for θ over the given domain

1. Solve for θ over the given domain

√3 sin 2θ = cos 2θ
Domain: 0o ≤ θ ≤ 360o

2. Hello, sweetG!

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

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

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

3. 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

4. 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 $\tan \theta$ function has a period of $180^{\circ}$ hence when you add $180^{\circ}$, you get the value for $\theta$ which will satisfy your equation. There are infinite solution hence questions usually provide a range (in this case $0^o \leq \theta \leq 360^o$).

5. Thanks Soroban and Air that really helped and now I'm a lot more clearer on how to do these types of questions