Question

Cot^4θ-4cot^2θ-5=0 solve for θ

Answer 1

$\cot^4\theta-4\cot^2\theta-5=(\cot^2\theta-5)(\cot^2\theta+1)=0$

$\cot^2\theta-5=0\implies\cot^2\theta=5\implies\cot\theta=\pm\sqrt5$

Recall that $\cot\theta=\cot(\theta+\pi)$, which is to say $\cot x$ has period $\pi$. This in turn means that $\cot\theta=0$ will have the same solutions as $\cot(\theta+n\pi)=0$ for any integer $n$. So the general solution to the first case is

$\cot\theta=\pm\sqrt5\implies\theta=\cot^{-1}(\pm\sqrt5)+n\pi$

where $n$ is any integer.

On the other hand,

$\cot^2\theta+1=0\implies\cot^2\theta=-1$

but $x^2\ge0$ for any value of $x$, so this equation has no (real) solutions for $\theta$.