469. $2 \sin ^{2} x-3 \sin x-2=0$.

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469. $2 \sin ^{2} x-3 \sin x-2=0$.
Solution. Let $\sin x=y$, we get the quadratic equation $2 y^{2}-3 y-2=0$. Here $a=2, b=-3, c=-2, \quad D=b^{2}-4 a c=(-3)^{2}-$ $-4 \cdot 2 \cdot(-2)=9+16=25, \sqrt{D}=5$. Therefore, $y_{1}=\frac{3-5}{4}=-\frac{1}{2}, \quad y_{2}=$ $=\frac{3+5}{4}=2$.

Thus, $y_{1}=-\frac{1}{2}$ and $y_{2}=2$, i.e., $\sin x=-\frac{1}{2}$ and $\sin x=2$. The solution of the equation $\sin x=-\frac{1^{2}}{2}$ is $x=(-1)^{k} \arcsin \left(-\frac{1}{2}\right)+\pi k=(-1)^{k+1} \times$ $\times \arcsin \frac{1}{2}+\pi k$, i.e., $x_{1}=(-1)^{k+1} \frac{\pi}{6}+\pi k, k \in Z$. The equation $\sin x=2$ has no solutions.