I need to prove or disprove there exists $\left(k_1, k_2, k_3, k_4\right) \in \mathbb{R}^4$ such that the following equation holds only if $\left(m_1, m_2\right)=\left(t_1, t_2\right)$ (where $t_1, t_2$ are two fixed numbers both from $\left.[0,1]\right)$ and not for any other values of $\left(m_1, m_2\right) \in[0,1]^2$ $$ \begin{aligned} & \frac{t_2}{2}-2\left(k_1+\pi k_2\left(-1+2 m_1\right)\right) \cos \left(2 \pi m_1\right)-2\left(k_2-\pi k_1\left(-1+2 m_1\right)\right) \sin \left(2 \pi m_1\right) \\ & =\frac{t_1}{2}-2\left(k_3+\pi k_4\left(-1+2 m_2\right)\right) \cos \left(2 \pi m_2\right)-2\left(k_4-\pi k_3\left(-1+2 m_2\right)\right) \sin \left(2 \pi m_2\right) \end{aligned} $$ Obviously, there exists $\left(k_1^*, k_2^*, k_3^*, k_4^*\right)=\boldsymbol{k}^*$ such that the following holds $$ \begin{aligned} & \frac{t_2}{2}-2\left(k_1^*+\pi k_2^*\left(-1+2 t_1\right)\right) \cos \left(2 \pi t_1\right)-2\left(k_2^*-\pi k_1^*\left(-1+2 t_1\right)\right) \sin \left(2 \pi t_1\right) \\ & =\frac{t_1}{2}-2\left(k_3^*+\pi k_4^*\left(-1+2 t_2\right)\right) \cos \left(2 \pi t_2\right)-2\left(k_4^*-\pi k_3^*\left(-1+2 t_2\right)\right) \sin \left(2 \pi t_2\right) \end{aligned} $$ The proof would be just to manipulate the expression to find one $k_i$ in terms of other free variables. However, substituting $\boldsymbol{k}^*$ into the equation, there may be other $\left(m_1, m_2\right)$ not equal to $\left(t_1, t_2\right)$ such that the following is true. $$ \begin{aligned} & \frac{t_2}{2}-2\left(k_1^*+\pi k_2^*\left(-1+2 m_1\right)\right) \cos \left(2 \pi m_1\right)-2\left(k_2^*-\pi k_1^*\left(-1+2 m_1\right)\right) \sin \left(2 \pi m_1\right) \\ & =\frac{t_1}{2}-2\left(k_3^*+\pi k_4^*\left(-1+2 m_2\right)\right) \cos \left(2 \pi m_2\right)-2\left(k_4^*-\pi k_3^*\left(-1+2 m_2\right)\right) \sin \left(2 \pi m_2\right) \end{aligned} $$ Hence, these $\boldsymbol{k}^*$ do not satisfy the proposition I have stated.
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