The Schrödinger equation in Physics is solved by different wave functions. The solution depends on the environment and the starting values. In one dimension and without regarding the time, the equation looks like $$-\frac{\hbar}{2m}*\partial_x^2 \Psi(x)+V(x)*\Psi(x)=E*\Psi(x).$$ $V(x)$ is an initial energy. In this simple case, we start with $V(x)=0$ on an interval $(0,L)$. Outside, it shall go against $\infty$ to both sides along $x$. Furthermore, the solution $\Psi$ shall have the value $$\Psi(0)=0\quad\land\quad \Psi(L)=0.$$
Then, the solution may follow two solution terms: $$\Psi=A*\sin(kx)+B*\cos(kx)\quad\lor\quad \Psi=A*e^{ikx}+B*e^{-ikx}$$ and both terms in the Schrödinger equation lead to the solution of the equation. $A,B$ are going to differ in both cases but solutions are going to be very similar: $$\Psi=\sqrt{\frac{2}{L}}*\sin\left ( \frac{\pi*x}{L} \right ) \qquad\lor \qquad\Psi=2*\sqrt{\frac{1}{2L}}*\sin\left ( \frac{\pi*x}{L} \right ) .$$
To find the coefficient in the solution, one must solve the condition that $$\int_{-\infty}^{\infty} |\Psi |^2 \,\mathrm{d}x = 1.$$
That does not make a difference since both solution terms only differ in the coefficient $i$ (or not $i$). As you can see, it does not help that we have the values $0$ as starting values on certain points, because $$0=0*i.$$
How to find a unique solution for this problem? What is needed? Where is the mistake?
It seems that you made an error solving for the constant factor. In the first case $\Psi_1(x) = A_1\sin(kx)+B_1\cos(kx)$, the boundary values for $\Psi$ tell us that $B_1=0$ and $k= \frac{n\pi}{L}$ for $n\in \mathbb{Z}$. So our solution using this form of the solution is: $$\Psi_1(x) = A_1\sin\left(\frac{n\pi x}{L}\right)$$ Using the other form of the solution, $\Psi(0) = 0$ tells us that $$\Psi_2(x) = A_2(e^{ikx}-e^{-ikx})=2A_2i\sin(kx)$$ The other boundary term gives, again, $k=\frac{n\pi}{L}$ for $n\in \mathbb{Z}$. In both situations, to obtain the value of $A$, you must normalize the wavefunction. Since you seem to be interested in the case $n=1$, I will set $n=1$. Notice the value of the following integral: $$\int_{[0,L]} \sin^2\left(\frac{\pi x}{L}\right)\;dx = \int_{[0,L]}\frac{1-\cos(2\pi x/L)}{2}\;dx = \frac{L}{2}$$ So for the first form of your wavefunction, we have: $$|A_1|^2\frac{L}{2}=1\Longrightarrow |A_1| = \sqrt{\frac{2}{L}}$$ For the second form, we have: $$4|A_2|^2\frac{L}{2} = 1\Longrightarrow |A_2| = \sqrt{\frac{1}{2L}}$$ You typically want the constant in front of $\sin(\pi x/L)$ to be real, so in the second case, let $A_2= \frac{1}{i}\sqrt{\frac{1}{2L}}$. Then we have in the first case: $$\Psi_1(x) = A_1\sin\left(\frac{\pi x}{L}\right) = \sqrt{\frac{2}{L}}\sin\left(\frac{\pi x}{L}\right)$$ In the second case, we have: $$\Psi_2(x) = 2A_2i\sin\left(\frac{\pi x}{L}\right)=2\frac{1}{i}\sqrt{\frac{1}{2L}}i\sin\left(\frac{\pi x}{L}\right)=\sqrt{\frac{2}{L}}\sin\left(\frac{\pi x}{L}\right)$$ The solutions are identical.