Find all solutions (0 - 2pi) for "tan2x^2 - 1 = 0 "

[estex198]

tan²2x - 1 = 0. Find all solutions from 0 to 2pi

1.) sqrt(tan²2x) = sqrt(1)
2.) tan2x = +- 1 (reason for two quadratic equations in #4)
3.) use double angle identity for tan and rationalize to form two quadratic equations:
4.) tan²x + 2tanx - 1 AND tan²x -2tanx -1
5.) now I use quadratic equation but I've come to a roadblock with the second equation in #4 because I'm getting irrational solutions: (1 +- sqrt(2))
6.) I'm trying to find exact answer in radians. I'm not even sure I'm getting a correct answer. I'm totally stumped here and need help before moving on.

 

[Prove It]

tan²2x - 1 = 0. Find all solutions from 0 to 2pi

1.) sqrt(tan²2x) = sqrt(1)
2.) tan2x = +- 1 (reason for two quadratic equations in #4)
3.) use double angle identity for tan and rationalize to form two quadratic equations:
4.) tan²x + 2tanx - 1 AND tan²x -2tanx -1
5.) now I use quadratic equation but I've come to a roadblock with the second equation in #4 because I'm getting irrational solutions: (1 +- sqrt(2))
6.) I'm trying to find exact answer in radians. I'm not even sure I'm getting a correct answer. I'm totally stumped here and need help before moving on.

From $\displaystyle \begin{align*} \tan{(2x)} = \pm 1 \end{align*}$ just remember $\displaystyle \begin{align*} \arctan{(-1)} = -\frac{\pi}{4}, \arctan{(1)} = \frac{\pi}{4} \end{align*}$ and the period of the tangent function is $\displaystyle \begin{align*} \pi \end{align*}$, giving

$\displaystyle \begin{align*} 2x &= \pm \frac{\pi}{4} + \pi n \textrm{ where } n \in \mathbf{Z} \\ x &= \pm \frac{\pi}{8} + \frac{\pi}{2} n \end{align*}$

and use this to get all the solutions in the region $\displaystyle \begin{align*} x \in [0, 2\pi] \end{align*}$.

 

[estex198]

For the first quadratic in #4 i have "-1 +- sqrt(2)" How can I find exact angles from these in radians??

 

[Prove It]

For the first quadratic in #4 i have "-1 +- sqrt(2)" How can I find exact angles from these in radians??

You don't. You solve the equation $\displaystyle \begin{align*} \tan{(2x)} = \pm 1 \end{align*}$ directly without resorting to a quadratic.

 

[estex198]

From $\displaystyle \begin{align*} \tan{(2x)} = \pm 1 \end{align*}$ just remember $\displaystyle \begin{align*} \arctan{(-1)} = -\frac{\pi}{4}, \arctan{(1)} = \frac{\pi}{4} \end{align*}$ and the period of the tangent function is $\displaystyle \begin{align*} \pi \end{align*}$, giving

$\displaystyle \begin{align*} 2x &= \pm \frac{\pi}{4} + \pi n \textrm{ where } n \in \mathbf{Z} \\ x &= \pm \frac{\pi}{8} + \frac{\pi}{2} n \end{align*}$

and use this to get all the solutions in the region $\displaystyle \begin{align*} x \in [0, 2\pi] \end{align*}$.

Thanks! You really made this easy... Since I'm trying to get better at trig equations, I hope you don't mind my next question... Is there is another way?

 

[Prove It]

I'm sure there probably are, but multiple angle identities exist to make solving equations simpler, not more difficult.

 

[MarkFL]

Thanks! You really made this easy... Since I'm trying to get better at trig equations, I hope you don't mind my next question... Is there is another way?

I would approach it just as has Prove It. I would express the solution a bit differently:

$$2x=\frac{\pi}{4}+k\frac{\pi}{2}=\frac{\pi}{4}(2k+1)$$ where $k$ is an arbitrary integer.

Hence:

$$x=\frac{\pi}{8}(2k+1)$$ and then let $k$ be all those values which place $x$ in the given domain.

 

[Prove It]

Having reviewed your work, you have made some mistakes. First of all, the double angle identity is $\displaystyle \begin{align*} \tan{(2x)} \equiv \frac{2\tan{(x)}}{1 - \tan^2{(x)}} \end{align*}$, and so using it in your problem gives

$\displaystyle \begin{align*} \tan^2{(2x)} &= 1 \\ \left[ \frac{2\tan{(x)}}{1 - \tan^2{(x)}} \right] ^2 &= 1 \\ \frac{4\tan^2{(x)}}{1 - 2\tan^2{(x)} + \tan^4{(x)}} &= 1 \\ 4\tan^2{(x)} &= 1 - 2\tan^2{(x)} + \tan^4{(x)} \\ 0 &= \tan^4{(x)} - 6\tan^2{(x)} + 1 \end{align*}$

While this is a quadratic equation (you will need to let $\displaystyle \begin{align*} X = \tan^2{(x)} \end{align*}$ to solve it) it is still an extremely difficult problem to try to solve...