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(D) Given $f(x) = e^{\sin x} + 2m\sin x + 1$. For $f(x)$ to be increasing,$f'(x) > 0$ for all $x \in \left( 0, \frac{\pi}{2} \right)$. $f'(x) = \cos x

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$\left[ -\frac{1}{2}, \infty \right)$

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(D) Given $f(x) = e^{\sin x} + 2m\sin x + 1$. For $f(x)$ to be increasing,$f'(x) > 0$ for all $x \in \left( 0, \frac{\pi}{2} \right)$. $f'(x) = \cos x \cdot e^{\sin x} + 2m \cos x = \cos x (e^{\sin x} + 2m)$. Since $x \in \left( 0, \frac{\pi}{2} \right)$,$\cos x > 0$. Thus,we require $e^{\sin x} + 2m > 0$ for all $x \in \left( 0, \frac{\pi}{2} \right)$. This implies $2m > -e^{\sin x}$,or $m > -\frac{e^{\sin x}}{2}$. As $x$ varies from $0$ to $\frac{\pi}{2}$,$\sin x$ varies from $0$ to $1$. Therefore,$e^{\sin x}$ varies from $e^0 = 1$ to $e^1 = e$. So,$-\frac{e^{\sin x}}{2}$ varies from $-\frac{1}{2}$ to $-\frac{e}{2}$. For the inequality $m > -\frac{e^{\sin x}}{2}$ to hold for all $x$,$m$ must be greater than or equal to the supremum of the set $\left\{ -\frac{e^{\sin x}}{2} : x \in \left( 0, \frac{\pi}{2} \right) \right\}$. The supremum of $-\frac{e^{\sin x}}{2}$ is $-\frac{1}{2}$. Thus,$m \ge -\frac{1}{2}$,which is $\left[ -\frac{1}{2}, \infty \right)$.

The complete set of values of $m$ for which the function $f(x) = e^{\sin x} + 2m\sin x + 1$ is increasing for all $x \in \left( 0, \frac{\pi}{2} \right)$ is:

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