$8.419 \sin x+\sqrt{2-\sin ^{2} x}+\sin x \cdot \sqrt{2-\sin ^{2} x}=3$.
Solution. Let $u=\sin x, v=\sqrt{2-\sin ^{2} x}$.

$$
\begin{aligned}
& \left\{\begin{array} { l } 
{ u + v + u v = 3 , } \\
{ u ^ { 2 } + v ^ { 2 } = 2 }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ u + v = 3 - u v , } \\
{ ( u + v ) ^ { 2 } = 2 + 2 u v }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
u+v=3-u v, \\
u^{2} v^{2}-8 u v+7=0
\end{array} \Leftrightarrow\right.\right.\right. \\
& \Leftrightarrow\left[\begin{array} { l } 
{ \{ \begin{array} { l } 
{ u + v = - 4 , } \\
{ u v = 7 , } \\
{ \{ u + v = 2 , } \\
{ u v = 1 }
\end{array} }
\end{array} \Leftrightarrow \left[\begin{array}{l}
\varnothing, \\
u=v=1
\end{array} \Leftrightarrow \sin x=1 \Leftrightarrow x=\frac{\pi}{2}+2 \pi k\right.\right.
\end{aligned}
$$

Answer: $\quad x=\frac{\pi}{2}+2 \pi k$.

$8.4202-\sin x=\frac{\sqrt{1+\sin x}-\sqrt{1-\sin x}}{\sqrt{1+\sin x}+\sqrt{1-\sin x}}$.

Solution. Note that $\sqrt{1+\sin x}-\sqrt{1-\sin x}=0$ only when $\sin x=0$. But $x=\pi k$ is not a solution to the original equation. Therefore, we can eliminate the irrationality in the denominator of the right-hand side without violating equivalence. Thus, $2-\sin x=\frac{(\sqrt{1+\sin x}-\sqrt{1-\sin x})^{2}}{2 \sin x} \Leftrightarrow\left\{\begin{array}{l}(1-\sin x)^{2}=\sqrt{1-\sin x} \cdot \sqrt{1+\sin x}, \\ \sin x \neq 0\end{array} \Leftrightarrow\right.$ $\Leftrightarrow\left\{\begin{array}{l}(1-\sin x)^{4}=(1-\sin x) \cdot(1+\sin x) \\ \sin x \neq 0 .\end{array}\right.$

1) $\sin x=1 \Leftrightarrow x=\frac{\pi}{2}+2 \pi k$
2) $\left\{\begin{array}{l}(1-\sin x)^{3}=1+\sin x, \\ \sin x \neq 0\end{array} \Leftrightarrow \sin ^{2} x-3 \sin x+4=0 \Leftrightarrow \varnothing\right.$.

Answer: $\quad x=\frac{\pi}{2}(4 k+1)$.

## $8.421 \sqrt[3]{2-\operatorname{tg} x}+\sqrt[3]{7+\operatorname{tg} x}=3$.

Solution. Raise both sides of the equation to the third power:

$9+3 \sqrt[3]{2-\operatorname{tg} x} \cdot \sqrt[3]{7+\operatorname{tg} x} \cdot(\sqrt[3]{2-\operatorname{tg} x}+\sqrt[3]{7+\operatorname{tg} x})=27 \Leftrightarrow \sqrt[3]{2-\operatorname{tg} x} \sqrt[3]{7+\operatorname{tg} x}=2 \Leftrightarrow$ $\Leftrightarrow(2-\operatorname{tg} x) \cdot(7+\operatorname{tg} x)=8 \Leftrightarrow \operatorname{tg}^{2} x+5 \operatorname{tg} x-6=0 \Leftrightarrow\left[\begin{array}{c}\operatorname{tg} x=-6 \\ \operatorname{tg} x=1 .\end{array}\right.$

Answer: $x_{1}=-\operatorname{arctg} 6+\pi k, x_{2}=\frac{\pi}{4}(4 k+1)$.

## $8.422 \sqrt{4 \cos ^{2} x+1}+\sqrt{4 \sin ^{2} x+3}=4$

Solution. Raise both sides of the equation to the second power:

$$
\begin{aligned}
& 8+2 \sqrt{4 \cos ^{2} x+1} \cdot \sqrt{4 \sin ^{2} x+3}=16 \Leftrightarrow\left(4 \cos ^{2} x+1\right)\left(4 \sin ^{2} x+3\right)=16 \Leftrightarrow \\
& \Leftrightarrow 16 \sin ^{2} x \cdot \cos ^{2} x+8 \cos ^{2} x-9=0 \Leftrightarrow 4 \sin ^{2} 2 x+4(1+\cos 2 x)-9=0 \Leftrightarrow \\
& \Leftrightarrow 4 \cos ^{2} 2 x-4 \cos 2 x+1=0 \Leftrightarrow(2 \cos 2 x-1)^{2}=0 \Leftrightarrow \cos 2 x=\frac{1}{2} \Leftrightarrow x= \pm \frac{\pi}{6}+\pi k
\end{aligned}
$$

Answer: $x=\frac{\pi}{6}(6 k \pm 1)$.