For $f(x)=\sin ^{2} x-\cos x$, and $0 \leq x \leq \pi$, find the value(s) of $x$ for each condition listed below. NOTE: Enter the exact answers, or round to 3 decimal places if needed. (a) Number of local maxima: Choose one $\nabla$ (b) Number of local minima: Choose one $r$ (c) Global maximum at $x=$ (d) Global minimum at $x=$

Step 1 ：First, we find the derivative of the function: $f'(x) = 2\sin{x}\cos{x} + \sin{x}$

Step 2 ：To find the critical points, we set $f'(x) = 0$: $2\sin{x}\cos{x} + \sin{x} = 0$

Step 3 ：Factor out \(\sin{x}\): $\sin{x}(2\cos{x} + 1) = 0$

Step 4 ：The critical points are $x = 0, \pi, \frac{2\pi}{3}$

Step 5 ：Next, we find the second derivative: $f''(x) = 2\cos^2{x} - 2\sin^2{x} - \cos{x}$

Step 6 ：Evaluate $f''(x)$ at the critical points:

Step 7 ：$f''(0) = 2 - 1 = 1 > 0$, so $x = 0$ is a local minimum

Step 8 ：$f''(\pi) = 2 - 1 = 1 > 0$, so $x = \pi$ is a local minimum

Step 9 ：$f''(\frac{2\pi}{3}) = 2\left(\frac{1}{4}\right) - 2\left(\frac{3}{4}\right) - \left(-\frac{1}{2}\right) = -1 < 0$, so $x = \frac{2\pi}{3}$ is a local maximum

Step 10 ：Since there is only one local maximum, it is also the global maximum. Similarly, the global minimum is the smaller of the two local minima.

Step 11 ：Thus, the global maximum is at $x = \frac{2\pi}{3}$ and the global minimum is at $x = 0$

Step 12 ：\(\boxed{\text{(a) } x = \frac{2\pi}{3}}\)

Step 13 ：\(\boxed{\text{(b) } x = 0, \pi}\)

Step 14 ：\(\boxed{\text{(c) } x = \frac{2\pi}{3}}\)

Step 15 ：\(\boxed{\text{(d) } x = 0}\)