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

estex198

New member
Feb 7, 2014
14
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

Well-known member
MHB Math Helper
Jan 26, 2012
1,404
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

New member
Feb 7, 2014
14
For the first quadratic in #4 i have "-1 +- sqrt(2)" How can I find exact angles from these in radians??
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,404
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

New member
Feb 7, 2014
14
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 dont mind my next question... Is there is another way?
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,404
I'm sure there probably are, but multiple angle identities exist to make solving equations simpler, not more difficult.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Thanks! You really made this easy... Since I'm trying to get better at trig equations, I hope you dont 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:

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

Hence:

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

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,404
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...