Solving trigonometric equation with unknown and restricted domain Given that $ \tan^2(\fracθ3) = 1$ and $θ\in [0, 4\pi]$ find θ.
I'm not sure how to progress with the restricted domain. Here's what I've got so far:
Solving for the domain $[0, 4\pi]$.
$$ \tan^2(\fracθ3) = 1$$
$$ \tan(\fracθ3) = 1$$
Since $ \tan^{-1}(1) = \frac\pi4$
$$ \fracθ3 = \frac\pi4$$
$$ θ = 3\pi/4$$
How should I deal with the domain to solve the equation?
 A: HINT:
$$\text{If }\tan^2x=\tan^2A$$
$x=n\pi\pm A$ where $n$ is any integer   (Please prove this)
You have already found $A\left(=\frac\theta3, \text{here}\right)=\frac\pi4$

Alternatively, use $$\cos2x=\frac{1-\tan^2x}{1+\tan^2x}$$  
and $$\cos y=\cos A\implies y=2m\pi\pm A$$ where $m$ is any integer
and another form of $y,$ in $\cos y=0$ is $y=(2r+1)\frac\pi2$ where $r$ is any integer
A: There's no reason not to change the problem to $ tan^2(\phi) = 1$ and $\phi\in [0, 12\pi]$.  It may be less confusing for you this way.  Then just remember to multiply your final answer by 3, since $\theta = 3 \phi$.
How many revolutions is $12 \cdot pi$?  Six right?
$tan(\phi) = \pm 1$
If you have a vector at $(x, y)$ at an angle of $\phi$ off the positive X axis, what is $tan(\phi)$?

You should be able to tell from basic trig that the tangent of the angle is $y/x$.  So, we are looking for the $\phi$ for which $y/x = \pm 1$, in other words, the x distance is the same as the y distance.
You should find 4 angles for which the x value is the same as the y value.  One of them is shown above, $\phi = pi / 4$.  But that is just 4 angles in one revolution.  Your range is six revolutions.  That's 24 total angles.
So your solution should include the six solutions:  $\phi = \frac \pi 4, \phi = \frac \pi 4 + 2\pi, ... \phi = \frac \pi 4 + 5 \cdot 2\pi$.
The other 24 - 6 = 18 solutions are from the six revolutions of the other 3 angles.
Don't forget to write your final answer in terms of $\theta$, not $\phi$.
A: Put ${\theta\over3}=:\alpha\in[0,{4\pi\over3}\bigr]$. Then $\tan^2\alpha=1$, whence $\tan\alpha\in\{-1,1\}$. Taking the restriction on $\alpha$ into account we  find $\alpha\in\bigl\{{\pi\over4}, {3\pi\over4},{5\pi\over4}\bigr\}$ as possible solutions. The solution set for $\theta$ is therefore given by $\bigl\{{3\pi\over4}, {9\pi\over4},{15\pi\over4}\bigr\}$.
