Determine the intervals in which $f:R \rightarrow R$, $f\left(x\right) = 2 \cos x + \sin^2 x$ is strictly increasing or strictly decreasing.
$f\left(x\right) = 2 \cos x + \sin^2 x$
$\therefore$ $f'\left(x\right) = -2 \sin x + 2 \sin x \cos x$
i.e. $f'\left(x\right) = -2 \sin x \left(1 - \cos x\right) $
i.e. $f'\left(x\right) = -2 \sin x \times 2 \sin^2 \left(\dfrac{x}{2}\right)$
i.e. $f'\left(x\right) = -4 \sin x \sin^2 \left(\dfrac{x}{2}\right)$ $\;\; \cdots$ (1)
Now, $\sin^2 \left(\dfrac{x}{2}\right) > 0 \;\; \forall \; x \in R$
When $f'\left(x\right) > 0$ $\implies$ $f\left(x\right)$ is increasing.
Now, from equation (1), $f'\left(x\right) > 0$ $\implies$ $\sin x < 0$
$\implies$ $- \pi < x < 0$
Generalizing, we have $\left(2 k - 1\right)\pi < x < 2 k \pi$
i.e. $f\left(x\right)$ is strictly increasing in the interval $\left(\left(2 k - 1\right)\pi, 2 k \pi\right), \;\; k \in Z$
When $f'\left(x\right) < 0$ $\implies$ $f\left(x\right)$ is decreasing.
Now, from equation (1), $f'\left(x\right) < 0$ $\implies$ $\sin x > 0$
$\implies$ $0 < x < \pi$
Generalizing, we have $2 k \pi < x < \left(2 k + 1\right) \pi$
i.e. $f\left(x\right)$ is strictly decreasing in the interval $\left(2 k \pi, \left(2 k + 1 \right) \pi \right), \;\; k \in Z$