How to solve the Sturm-Liouville Problem $y''+\lambda y=0; y(0)-2y'(0)=0, y'(1)=0$

Question

I've been able to show that the cases $\lambda =0, \lambda < 0,$ do not have any eigenvalues; however, I'm stuck solving for $\lambda > 0.$ Here is where I get stuck:
let $\lambda = k^2, k > 0.$ The general solution is $$y(x)=c_1\cos(kx)+c_2\sin(kx)$$ And the derivate of $y(x) $ is $$y'(x)=-kc_1\sin(kx)+kc_2\cos(kx)$$
$y(0)-2y'(0) = 0$ simplifies to $$c_1-2kc_2=0 \implies c_2 = \frac{1}{2k}c_1$$ Then by substituting $c_2$ into $y'(1)=0$ $$kc_1\left(-\sin(k)+\frac{1}{2k}\cos(k) \right)=0$$ $c_1 \neq0$ $$-\sin(k)+\frac{1}{2k}\cos(k)=0$$ by solving for $\lambda$ we get $$\frac{1}{2\sqrt{\lambda}}=\tan(\sqrt{\lambda})$$ I know the real answer is $\frac{1}{2}\sqrt{\lambda}=\tan(\sqrt{\lambda})$, which doesn’t match the answer that I got. So I'm not sure where I'm making a mistake.