# Solve tanx=cotx for all solutions [0, 2pi)?

Oct 17, 2016

$x \in \left\{\frac{\pi}{4} , \frac{3 \pi}{4} , \frac{5 \pi}{4} , \frac{7 \pi}{4}\right\}$

#### Explanation:

Note that the initial presence of $\tan \left(x\right) = \sin \frac{x}{\cos} \left(x\right)$ and $\cot \left(x\right) = \cos \frac{x}{\sin} \left(x\right)$ implies we must have $\sin \left(x\right) \ne 0$ and $\cos \left(x\right) \ne 0$. With that:

$\tan \left(x\right) = \cot \left(x\right)$

$\implies \sin \frac{x}{\cos} \left(x\right) = \cos \frac{x}{\sin} \left(x\right)$

$\implies \sin \frac{x}{\cos} \left(x\right) \cdot \sin \left(x\right) \cos \left(x\right) = \cos \frac{x}{\sin} \left(x\right) \cdot \sin \left(x\right) \cos \left(x\right)$

$\implies {\sin}^{2} \left(x\right) = {\cos}^{2} \left(x\right)$

$\implies \sin \left(x\right) = \pm \cos \left(x\right)$

If we examine a unit circle, we find that this equality holds at $x = \frac{\pi}{4} + n \frac{\pi}{2} , n \in \mathbb{Z}$. Thus, we must only find which values of $n$ cause $x$ to lay within the interval $\left[0 , 2 \pi\right)$ Testing, we find that

$\frac{\pi}{4} + n \frac{\pi}{2} \in \left[0 , 2 \pi\right)$ for $n \in \left\{0 , 1 , 2 , 3\right\}$

Substituting those in, we get our answers:

$x \in \left\{\frac{\pi}{4} , \frac{3 \pi}{4} , \frac{5 \pi}{4} , \frac{7 \pi}{4}\right\}$