Let $\mathbf{A}\in\mathbb{C}^{m\times m}$, and $\mathbf{B}\in\mathbb{H}^{m\times m}$ be an $m$-dimension Hermitian matrix, solve $\theta$ that satisfies the condition \begin{equation*} e^{\jmath\theta}\mathbf{A}+e^{-\jmath\theta}\mathbf{A}^H+\mathbf{B}\succeq0, \end{equation*} where $\jmath=\sqrt{-1}$.
Thank you so much.
(Too long for a comment.)
The solution $\theta$ (which is presumably real) may not exist. This occurs, e.g., when $A=0$ and $B=-I$. I don't know any sufficient existential condition that isn't too demanding. (It is easy to cook up a sufficient condition, such as $B\succeq2\|A\|_2I$, but that would make every $\theta$ a solution.)
Here is a reformulation of the problem that may or may not be useful. Let $c>0$. Rewrite the inequality as $$ \left(\frac1c e^{\jmath\theta}A + cI\right)\left(\frac1c e^{-\jmath\theta}A^H + cI\right) \succeq \frac{1}{c^2}AA^H + c^2I - B.\tag{1} $$ Choose a sufficiently large $c$ so that the RHS is positive definite. Let $P$ be the positive definite square root of the RHS. Then $(1)$ is equivalent to $$ \sigma_\min\left(\frac1c e^{\jmath\theta}P^{-1}A + cP^{-1}\right)\ge1.\tag{2} $$ So, the problem boils down to maximising the minimum singular value of $\frac1c e^{\jmath\theta}P^{-1}A + cP^{-1}$ over $\theta\in[0,2\pi]$. (You don't need to find the global maximiser. Any $\theta$ that satisfies $(2)$ suffices as a solution to the original problem.)