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

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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.
 
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estex198 said:
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 $\displaystyle \begin{align*} x \in [0, 2\pi] \end{align*}$.
 
For the first quadratic in #4 i have "-1 +- sqrt(2)" How can I find exact angles from these in radians??
 
estex198 said:
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.
 
Prove It said:
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?
 
I'm sure there probably are, but multiple angle identities exist to make solving equations simpler, not more difficult.
 
estex198 said:
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.
 
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...
 
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