Max and min of $f(x)=\sin^2{x}+\cos{x}+2$. My first try was to look for when $\cos{x}$ and $\sin{x}$ attain their min and max respectively, which is easy using the unit circle. So the minimum of $\sin{x}+\cos{x}$ has maximum at $\sqrt{2}$ and minimum at $-\sqrt{2}.$ But if the sine-term is squared, how do I find the minimum of $\sin^2{x}+\cos{x}$?
I then tried the differentiation method:
$$f'(x)=2\sin{x}\cos{x}-\sin{x}=(2\cos{x}-1)\sin{x}=0,$$
which in the interval $(0,2\pi)$, gave me the the roots $\left(\frac{\pi}{2},\frac{3\pi}{2},\frac{\pi}{3},\frac{5\pi}{3}\right)$. So
$$f(\pi/2)=3, \quad f(3\pi/2)=3, \quad f(\pi/3)=\frac{13}{4}, \quad f(5\pi/3)=\frac{13}{4}.$$
So $f_{\text{max}}=\frac{13}{4}$ and $f_{\text{min}}=3.$ But the minimum should be $1$.
 A: Credit to @iamwhoiam for suggesting this solution.
There's nothing wrong with your method - you just missed a root. When you find the roots of the first derivative, either $2 \cos x - 1$ or $\sin x$ should equal $0$, so we have:
First case: 
$2 \cos x - 1 = 0$, $\cos x = \frac{1}{2}$.
Since $\cos(x)$ is an even function, $\cos(x) = \cos(-x)$. Since we know that one solution is $x = \frac{\pi}{3}$, $x$ can also be $\frac{5\pi}{3}$.
Second case:
$\sin x = 0, \pi$. Using the properties of $\sin x$ we know these are the only solutions in the range $[0,2\pi)$.
Therefore the solutions in the range $[0, 2\pi)$ are: $x = 0, \frac{\pi}{3}, \frac{5\pi}{3}, \pi$. Substituting each in the original equation gives $3, \frac{13}{4}$, $\frac{13}{4}$ and $1$ respectively. The minimum of these critical points is $1$, which is the minimum value of the original function.
A: Hint: $\sin^2(x)=1-\cos^2(x)$. You will get a quadratic equation in $u=\cos(x)$.
A: HINT: write $f(x)$ as
$$f(x)=1-\cos(x)^2+\cos(x)+2 \\=-\cos(x)^2+\cos(x)+3\\=-\left(\cos(x)^2-\cos(x)+\frac{1}{4}\right)+\frac{13}{4}\\=-\left(\cos(x)-\frac{1}{2}\right)^2+\frac{13}{4}$$
A: Set $t=\cos x$. Then
$$\sin^2x+\cos x+2=1-\cos^2x+\cos x+2=3+t-t^2$$
is a quadratic  polynomial $ P(t)$, which has a global maximum for $t=\dfrac12$, and this maximum is equal  $$P\Bigl(\dfrac12\Bigr)=\dfrac{13}4.$$
Now we have to find the minimum of $P(t)$ on the interval $[-1,1]$. As this function is symmetric w.r.t. $\;x=\dfrac12$, the minimum is attained at $t=-1$, and it is equal to 
$\;P(-1)=1.$
A: As per your work
$$f'(x)=2\sin{x}\cos{x}-\sin{x}=(2\cos{x}-1)\sin{x}=0,$$
     now either  $(2\cos{x}-1)=0$ then ${x}={\pi/3,5\pi/3}$
     or  $\sin{x}=0$     then ${x}={\pi}$
for ${x}={\pi}$  value of f(${\pi}$)  =1 which is minimum value of function.  
