= Solution
{c}
For $\omega\ne0$, substitution of the <separated solutions> into the <heat equations> gives
$$
G_i''+\omega^2\alpha_i^2G_i=0,\qquad
\alpha_i=\sqrt{\frac{\rho_ic_i}{k_i}}.
$$
The zero exterior <boundary values> are incorporated by
$$
G_1(x)=A\sin[\omega\alpha_1(x+L)],\qquad
G_2(x)=B\sin[\omega\alpha_2(L-x)].
$$
To obtain a <eigenvalue equation for heat conduction in two joined slabs>, write $u=\omega\alpha_1L$ and $v=\omega\alpha_2L$. Continuity of <temperature> and <heat flux> at zero gives the homogeneous <linear system>
$$
A\sin u-B\sin v=0,\qquad
k_1\alpha_1A\cos u+k_2\alpha_2B\cos v=0.
$$
A nonzero amplitude pair exists precisely when its <determinant> vanishes, equivalently
$$
\boxed{D(\omega):=k_1\alpha_1\cos u\sin v+k_2\alpha_2\sin u\cos v=0,\qquad\omega\ne0.}
$$
If $\cos u\cos v\ne0$, division by $k_1k_2\alpha_1\alpha_2\cos u\cos v$ gives the printed <transcendental equation>
$$
\boxed{\frac{\tan(\omega\alpha_1L)}{\alpha_1k_1}
+\frac{\tan(\omega\alpha_2L)}{\alpha_2k_2}=0.}
$$
\b[The printed necessity claim needs the qualification that both <tangents> are defined.] If just one of $\cos u,\cos v$ vanishes, $D(\omega)\ne0$. If both <cosines> vanish, however, $D(\omega)=0$ and legitimate <normal modes> remain. Their <temperature> at the <thermal interface> need not be zero; the <temperature> condition fixes the amplitude ratio, while each <heat flux> is zero there.
For a concrete counterexample, take identical slabs: $\rho_1=\rho_2=\rho$, $c_1=c_2=c$, $k_1=k_2=k$, and $\alpha=\sqrt{\rho c/k}$. Then
$$
\boxed{T_1(x,t)=T_2(x,t)=e^{-\omega^2t}\cos\frac{\pi x}{2L},\qquad
\omega=\frac{\pi}{2\alpha L},}
$$
with each formula restricted to its own slab, satisfies both <heat equations> and all four <boundary conditions>. But $\omega\alpha L=\pi/2$, so neither <tangent> is defined. This directly disproves the unqualified printed assertion. The <determinant> condition retains these missing <normal modes>.
Finally, $\omega=0$ must be checked before dividing by it. Then $G_1=A(x+L)$, $G_2=B(L-x)$, and interface continuity gives $A=B$, while <heat flux> continuity gives $k_1A=-k_2B$. Positivity of the <thermal conductivities> forces $A=B=0$. Thus $\boxed{\omega=0\text{ gives only the trivial solution}.}$ In the <determinant> formula, its apparent zero at $\omega=0$ is spurious because the chosen sine basis degenerates there.
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