Sturm Liouville Problem Transcendental Equation I am trying to solve the following problem

$$ X''(x)+ \lambda X(x)=0$$
$$ X'(0)+2X(0)=0$$
$$ X'(1)=0$$
Show that
$$ \tan\left( \sqrt{ \lambda } \right)= -2/ \sqrt{ \lambda } $$
With the eigenfunctions
$$ X(x) = A\cos( \sqrt{ \lambda } (x-1)) $$

Lambda is positive real number and $x$ is between $0$ and $1$ but not  exactly $1$ or $0$.
 A: The general solution is given by $$X(x)=C_1\cos(x\sqrt{\lambda})+C_2\sin(x\sqrt{\lambda})$$
for arbitrary constants $C_1$ and $C_2$. Then $X'(x)=-C_1\sqrt{\lambda}\sin(x\sqrt{\lambda})+C_2\sqrt{\lambda}\cos(x\sqrt{\lambda})$, so we have
$$X'(0)+2X(0)=C_2\sqrt{\lambda}+2C_1=0\implies\space C_1=-\frac{\sqrt{\lambda}}{2}C_2$$
and
$$X'(1)=-C_1\sqrt{\lambda}\sin(\sqrt{\lambda})+C_2\sqrt{\lambda}\cos(\sqrt\lambda)$$
$$=\frac{\lambda}{2}C_2\sin(\sqrt{\lambda})+C_2\sqrt{\lambda}\cos(\sqrt{\lambda})=0$$
which gives $\tan(\sqrt{\lambda})=-\frac{2}{\sqrt{\lambda}}$ provided $\cos(\sqrt{\lambda})\neq 0.$ Then $$C_2=-\frac{2}{\sqrt{\lambda}}C_1=C_1\tan(\sqrt{\lambda})$$
So the eigenfunctions are given by
$$X(x)=C_1\cos(x\sqrt{\lambda})+C_2\sin(x\sqrt{\lambda})$$
$$C_1\big(\cos(x\sqrt{\lambda})+\tan(\sqrt{\lambda})\sin(x\sqrt{\lambda})\big)$$
$$=C_{1}\big(\frac{\cos(x\sqrt{\lambda})\cos(\sqrt{\lambda})+\sin(\sqrt{\lambda})\sin(x\sqrt{\lambda})}{\cos(\sqrt{\lambda})}\big)$$
$$=A\big(\cos(x\sqrt{\lambda})\cos(\sqrt{\lambda})+\sin(x\sqrt{\lambda})\sin(\sqrt{\lambda})\big)=A\cos(x\sqrt{\lambda}-\sqrt{\lambda})=A\cos(\sqrt{\lambda}(x-1))$$
where we have used the angle addition formula $\cos(\alpha-\beta)=\cos(\alpha)\cos(\beta)+\sin(\alpha)\sin(\beta)$ and $A=\frac{C_1}{\cos(\sqrt{\lambda})}$ is a constant.
