Minimize the expression

at $x=\frac{-\pi}{6}$

let $P=sinx-cos^{2}x-1$

using

$\frac{dP}{dx}=0$ , we get

$cosx(1+2sinx)=0$

$x=\frac{\pi}{2}$ or $x=n \pi+(-1)^{n}(\frac{-\pi}{6})$

clearly at $\frac{-\pi}{6}$ , $P$ attains minimum value

$\sin x-\cos^2 x-1=\sin x-(1-\sin^2 x)-1$

$=\left(\sin x+\frac{1}{2}\right)^2-\frac{9}{4}\ge -\frac{9}{4}$

with equality iff $\sin x=-\frac{1}{2}$

$\iff x=(-1)^m\arcsin\left(-\frac{1}{2} \right)+m\pi=\frac{11\pi(-1)^m}{6}+m\pi,\, m\in\Bbb Z$

Here I used (with $|a|\le 1$ ):

$\sin x=a\iff x=(-1)^m\arcsin a + m\pi, m\in\Bbb Z$

We also have (with $|a|\le 1$ ):

$\cos x=a\iff x=\pm \arccos a+2m\pi, m\in\Bbb Z$

Also (with any $a\in\Bbb R$ ):

$\tan x=a\iff x=\arctan a + m\pi, m\in\Bbb Z$

$\cot x=a\iff x=\text{arccot}\, a+m\pi, m\in\Bbb Z$