The function f(x) = sin x (1 + cos x) is maxi | vedclass

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(C) Given $f(x) = \sin x (1 + \cos x) = \sin x + \sin x \cos x = \sin x + \frac{1}{2} \sin 2x$. Find the first derivative: $f'(x) = \cos x + \cos 2x$.

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$x = \pi / 3$

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(C) Given $f(x) = \sin x (1 + \cos x) = \sin x + \sin x \cos x = \sin x + \frac{1}{2} \sin 2x$. Find the first derivative: $f'(x) = \cos x + \cos 2x$. Set $f'(x) = 0$ for critical points: $\cos x + (2 \cos^2 x - 1) = 0$. Rearranging gives $2 \cos^2 x + \cos x - 1 = 0$. Factoring the quadratic: $(2 \cos x - 1)(\cos x + 1) = 0$. This gives $\cos x = 1/2$ or $\cos x = -1$. For $x \in [0, \pi]$,$\cos x = 1/2 \implies x = \pi / 3$ and $\cos x = -1 \implies x = \pi$. Find the second derivative: $f''(x) = -\sin x - 2 \sin 2x$. At $x = \pi / 3$: $f''(\pi / 3) = -\sin(\pi / 3) - 2 \sin(2\pi / 3) = -\frac{\sqrt{3}}{2} - 2(\frac{\sqrt{3}}{2}) = -\frac{3\sqrt{3}}{2} Since $f''(\pi / 3) < 0$,the function has a local maximum at $x = \pi / 3$.

The function $f(x) = \sin x (1 + \cos x)$ is maximum at $x = \dots$

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