Find the least value for $\sin x - \cos^2 x -1$ 
Find all the values of $x$ for which the function $y = \sin x - \cos^2 x -1$ assumes the least value. What is that value?

At first I found the first derivative to be $y' = \cos x + 2 \sin x \cos x$. 
Critical point $0$, $-π/6$ (principal)
$y'' = - \sin x + 2(\cos 2x)$
Then substitution of $x$ by critical points I found minima. But my answer is incorrect. 
Correct minimum value is $-9/4$
 A: $$f(x)=\sin^2x+\sin{x}-2=\left(\sin{x}+\frac{1}{2}\right)^2-2.25\geq-2.25.$$
The equality occurs for $\sin{x}=-\frac{1}{2},$ which says that $-2.25$ is a minimal value.
A: Yes your way is correct indeed for $x=-\pi/6$ we have
$$f(x)=-\frac12-\frac34-1=-\frac94$$
As an alternative, recall that $\cos^2 x=1-\sin^2 x$ and therefore we have
$$f(x)=\sin x -\cos^2 x-1=\sin^2 x+\sin x-2$$
then set $t=\sin x$ and consider the parabola (concave up)
$$g(t)=t^2+t-2 \implies g'(t)=2t+1=0 \implies t=-\frac 12$$
A: The critical points in $[-\pi,\pi]$ are those for which
$$
\cos x(1+2\sin x)=0
$$
which means $x=\pi/2$, $x=-\pi/2$, $x=-\pi/6$ and $x=-5\pi/6$.
Note that
$$
y''=-\sin x+2\cos^2x-2\sin^2x
$$
Since
$$
y''(\pi/2)=-1-2=-3,
\quad
y''(-\pi/2)=1-2=-1,
\\
y''(-\pi/6)=\frac{1}{2}+2\frac{3}{4}-2\frac{1}{4}=\frac{3}{2}=y''(-5\pi/6)
$$
the points of local minimum are $-\pi/6$ and $-5\pi/6$.
Just compute
$$
y(-\pi/6)=-\frac{1}{2}-\frac{3}{4}-1=-\frac{9}{4}
$$
The value at $-5\pi/6$ is the same.

Simpler: write
$$
y=\sin x-1+\sin^2x-1=\sin^2x+\sin x-2=
\left(\sin x+\frac{1}{2}\right)^{\!2}-\frac{9}{4}
$$
Obviously, the minimum value is where $\sin x=-\frac{1}{2}$ and is $-9/4$.
The maximum is where $\sin x=1$.
