Unable to solve $y''+\lambda y =0$ I wish to find the eigenvalues and eigenfunctions of the following, but am unable to and further don't know what I am doing wrong at all

$y''+\lambda y =0$ where $y(0)=0$, $y'(1)+y(1)=0$

My attempt

Let $y=e^{\mu x}$, then $y''=\mu ^2 e^{\mu x}$. By substitution into the original equation we find that $\mu = \pm i \sqrt{\lambda}$. Thus $y=C_1 e^{i \sqrt{\lambda}x}+C_2 e^{-i \sqrt{\lambda}x}$. Note $C_1,C_2$ are constant coefficients.
Then, $y(0)=C_1+C_2=0$ hence $C_1=-C_2$ and vice versa using the first boundary condition. The second requires a first order differential of $y$ so, $y'=C_1i \sqrt{\lambda}e^{i \sqrt{\lambda}x}-C_2 i \sqrt{\lambda} e ^{-i \sqrt{\lambda} x}$. So the second boundary condition gives us,
$$C_1 e^{i \sqrt{\lambda}}+C_2 e^{-i \sqrt{\lambda}}+C_1i \sqrt{\lambda}e^{i \sqrt{\lambda}}-C_2 i \sqrt{\lambda} e ^{-i \sqrt{\lambda}}=0$$
Rearranging and using the sustitutiton $C_1=-C_2$, we obtain (I have used $z$ to express $1+i \sqrt{\lambda}$),
$$C_1(ze^{i \sqrt{\lambda}}-\bar z e^{-i \sqrt{\lambda}})=0$$

Here's where I am stuck. Simply, this helps me in no way to find what $C_1$ is. If I find $C_1$ or $C_2$, I can find the other and it's all good. What is more confusing, is that if I use $e^{ix}=\cos{x}+i\sin{x}$, it still gives me nothing since I will need $\sqrt{\lambda}$ such that $\sin{\sqrt{\lambda}}=0$ and $\cos{\sqrt{\lambda}}=0$ which clearly does not exist(I drew a graph of both sine and cosine to make sure; there is clearly no point that both of the equals zero simultaneously in the $0$ to $2 \pi$ interval).
Well, what am I doing wrong? Any arithmetic mistakes? Or does this involves some ingenious math-trick to solve?
 A: The eigenvalues are those values of $\lambda$ that allow a non-zero solution, just as the eigenvalues of a matrix A are those values which allow $Av=\lambda v$ for nonzero $v$.
So the eigenvalues $\lambda$ are the solutions, using Claude's answer, of $$\sin(\sqrt{\lambda})+\sqrt\lambda\cos(\sqrt\lambda)=0$$
Then $c_2$ doesn't have to be zero, and you get a non-zero eigenfunction $y(x)=\sin\sqrt\lambda x$.
If you plot that function of $\lambda$, you will see it has many zeros; I don't think they are easily expressed in simple functions.
A: The general solution of $y''(x)+\lambda\, y(x)=0$ is given by $$y(x)=c_1 \cos \left(\sqrt{\lambda } x\right)+c_2 \sin \left(\sqrt{\lambda } x\right)$$ Applying the first condition $$y(0)=0\implies c_1=0$$ So, what is left is $y(x)=c_2 \sin \left(\sqrt{\lambda } x\right)$. Now, the second condition $$y(1)+y'(1)=0 \implies c_2 \left(\sin \left(\sqrt{\lambda }\right)+\sqrt{\lambda } \cos
   \left(\sqrt{\lambda }\right)\right)=0$$ Since $\lambda$ is given, I do not see any other possibility beside $c_2=0$ which makes $y(x)=0$.
Edit
After Michael's answer.
The zeros of equation $\sin(x)+x\, \cos(x)=0$ cannot be expressed explicitely but we can notice that they are close to $(2k+1)\frac \pi 2$.
They can be approximated using $[1,1]$ Padé approximants according to $$x_k=  (2 k+1)\frac{\pi}{2} \Big(1+\frac{4}{6+ (2 k+1)^2\pi ^2}\Big)$$ Some values are reproduced below 
$$\left(
\begin{array}{ccc}
k & approx & exact \\
 0 & 1.96672 & 2.02876 \\
 1 & 4.91117 & 4.91318 \\
 2 & 7.97828 & 7.97828 \\
 3 & 11.0854 & 11.0855 \\
 4 & 14.2074 & 14.2074 \\
 5 & 17.3363 & 17.3364 \\
 6 & 20.4691 & 20.4692 \\
 7 & 23.6043 & 23.6043 \\
 8 & 26.7409 & 26.7409 \\
\end{array}
\right)$$
