If the function y = sin x(1 + cos x) is defin | vedclass

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(C) Given the function $y = \sin x + \sin x \cos x = \sin x + \frac{1}{2} \sin 2x$. To find the interval where $y$ is strictly increasing,we find the

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$\left(-\frac{\pi}{3}, \frac{\pi}{3}\right)$

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(C) Given the function $y = \sin x + \sin x \cos x = \sin x + \frac{1}{2} \sin 2x$. To find the interval where $y$ is strictly increasing,we find the derivative $\frac{dy}{dx}$. $\frac{dy}{dx} = \cos x + \cos 2x$. For $y$ to be strictly increasing,we require $\frac{dy}{dx} > 0$. $\cos x + (2 \cos^2 x - 1) > 0$. Let $t = \cos x$,then $2t^2 + t - 1 > 0$. Factoring the quadratic: $(2t - 1)(t + 1) > 0$. Since $t = \cos x$ and $x \in [-\pi, \pi]$,we know $-1 \le t \le 1$. For $(2t - 1)(t + 1) > 0$,we must have $t > \frac{1}{2}$ (since $t+1$ is always $\ge 0$ and $t 
eq -1$). Thus,$\cos x > \frac{1}{2}$. In the interval $[-\pi, \pi]$,$\cos x > \frac{1}{2}$ implies $x \in \left(-\frac{\pi}{3}, \frac{\pi}{3}ight)$. Therefore,the function is strictly increasing in the interval $\left(-\frac{\pi}{3}, \frac{\pi}{3}ight)$.

If the function $y = \sin x(1 + \cos x)$ is defined in the interval $[-\pi, \pi]$,then $y$ is strictly increasing in the interval

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