If the function f(x) = sin x - cos^2 x is def | vedclass

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(D) Given $f(x) = \sin x - \cos^2 x = \sin x - (1 - \sin^2 x) = \sin^2 x + \sin x - 1$. Let $t = \sin x$. Since $x \in [-\pi, \pi]$,$t \in [-1, 1]$. T

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$(-\frac{5\pi}{6}, -\frac{\pi}{2}) \cup (-\frac{\pi}{6}, \frac{\pi}{2})$

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(D) Given $f(x) = \sin x - \cos^2 x = \sin x - (1 - \sin^2 x) = \sin^2 x + \sin x - 1$. Let $t = \sin x$. Since $x \in [-\pi, \pi]$,$t \in [-1, 1]$. Then $g(t) = t^2 + t - 1$. For $f(x)$ to be strictly increasing,we need $f'(x) > 0$. $f'(x) = \cos x + 2 \cos x \sin x = \cos x(1 + 2 \sin x)$. Setting $f'(x) > 0$: Case $1$: $\cos x > 0$ and $1 + 2 \sin x > 0$. $\cos x > 0 \implies x \in (-\frac{\pi}{2}, \frac{\pi}{2})$. $1 + 2 \sin x > 0 \implies \sin x > -\frac{1}{2} \implies x \in (-\frac{\pi}{6}, \frac{7\pi}{6})$. Intersection: $x \in (-\frac{\pi}{6}, \frac{\pi}{2})$. Case $2$: $\cos x $\cos x $1 + 2 \sin x Intersection: $x \in (-\frac{5\pi}{6}, -\frac{\pi}{2})$. Combining both cases,$f(x)$ is strictly increasing in $(-\frac{5\pi}{6}, -\frac{\pi}{2}) \cup (-\frac{\pi}{6}, \frac{\pi}{2})$. Thus,the correct option is $D$.

If the function $f(x) = \sin x - \cos^2 x$ is defined on the interval $[-\pi, \pi]$,then $f$ is strictly increasing in the interval

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