5

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?

rschwieb
  • 153,510
moss
  • 51
  • 2
    Note, squares admit negative solutions, and the function is periodic as well. So find all solutions and then check which of those are in the interval you want. – Macavity Sep 29 '13 at 05:07

3 Answers3

3

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

lab bhattacharjee
  • 274,582
1

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 $\bigl\{{3\pi\over4}, {9\pi\over4},{15\pi\over4}\bigr\}$.

Christian Blatter
  • 226,825
1

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 $tan(\phi)$?

Tangent vs Angle

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 $\phi$.

DanielV
  • 23,556