$8.410 \sqrt{3} \sin t-\sqrt{2 \sin ^{2} t-\sin 2 t+3 \cos ^{2} t}=0$.
## Solution.

$$
\begin{aligned}
& \sqrt{3} \sin t=\sqrt{2 \sin ^{2} t-\sin 2 t+3 \cos ^{2} t} \Leftrightarrow \\
& \Leftrightarrow\left\{\begin{array} { l } 
{ \sin t \geq 0 , } \\
{ 3 \sin ^{2} t = 2 \sin ^{2} t - \sin 2 t + 3 \cos ^{2} t }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
\sin t \geq 0, \\
\sin ^{2} t + 2 \sin t \cdot \cos t - 3 \cos ^{2} t = 0
\end{array} \Leftrightarrow\right.\right. \\
& \Leftrightarrow\left\{\begin{array} { l } 
{ \sin t \geq 0 , } \\
{ \sin t = - \cos t \pm \sqrt { \cos ^{2} t + 3 \cos ^{2} t } }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ \sin t \geq 0 , } \\
{ [ \begin{array} { l } 
{ \sin t = \cos t , } \\
{ \sin t = - 3 \cos t }
\end{array} }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
\sin t \geq 0, \\
{\left[\begin{array}{l}
\tan t=1, \\
\tan t=-3
\end{array}\right.}
\end{array} \Leftrightarrow\right.\right.\right.
\end{aligned}
$$

$\Leftrightarrow\left[\begin{array}{l}t=\frac{\pi}{4}+2 k \pi \\ t=-\arctan 3+\pi(2 k+1)\end{array}\right.$

Answer: $\quad t_{1}=\frac{\pi}{4}+2 k \pi, t_{2}=-\arctan 3+\pi(2 k+1)$.

## $8.411 \sqrt[3]{\sin ^{2} x}+\sqrt[3]{\cos ^{2} x}=\sqrt[3]{4}$

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

$$
\begin{aligned}
& \sin ^{2} x+\cos ^{2} x+3 \sqrt[3]{\sin ^{2} x \cdot \cos ^{2} x} \cdot\left(\sqrt[3]{\sin ^{2} x}+\sqrt[3]{\cos ^{2} x}\right)=4 \Leftrightarrow \\
& \Leftrightarrow \sqrt[3]{\sin ^{2} x \cdot \cos ^{2} x} \cdot \sqrt[3]{4}=1 \Leftrightarrow 4 \sin ^{2} x \cdot \cos ^{2} x=1 \Leftrightarrow \sin ^{2} 2 x=1 \Leftrightarrow \\
& \Leftrightarrow \cos ^{2} 2 x=0 \Leftrightarrow \cos 2 x=0 \Leftrightarrow x=\frac{\pi}{4}(2 k+1)
\end{aligned}
$$

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

## $8.4122 \sin ^{2} t=\sqrt{\sin ^{2} t-16 \sin ^{2} t \cdot \cos ^{2} t \cdot \cos ^{2} 2 t+\cos ^{2} t}$.

Solution. The original equation is equivalent to the following:

$$
\begin{aligned}
& 2 \sin ^{2} t=\sqrt{1-4 \sin ^{2} 2 t \cdot \cos ^{2} 2 t} \Leftrightarrow 2 \sin ^{2} t=\sqrt{1-\sin ^{2} 4 t} \Leftrightarrow \\
& \Leftrightarrow 2 \sin ^{2} t=|\cos 4 t| \Leftrightarrow 1-\cos 2 t=\left|2 \cos ^{2} 2 t-1\right| \Leftrightarrow
\end{aligned}
$$

$$
\Leftrightarrow\left[\begin{array} { l } 
{ 1 - \cos 2 t = 2 \cos ^{2} 2 t - 1 , } \\
{ 1 - \cos 2 t = 1 - 2 \cos ^{2} 2 t }
\end{array} \Leftrightarrow \left[\begin{array} { l } 
{ \cos 2 t = \frac { - 1 + \sqrt { 17 } } { 4 } , } \\
{ \cos 2 t = 0 , } \\
{ \cos 2 t = \frac { 1 } { 2 } }
\end{array} \Leftrightarrow \left[\begin{array}{l}
2 t= \pm \arccos \frac{\sqrt{17}-1}{4}+2 k \pi \\
2 t=\frac{\pi}{2}+\pi k \\
2 t= \pm \frac{\pi}{3}+2 \pi k
\end{array}\right.\right.\right.
$$

Answer: $t_{1}= \pm \frac{1}{2} \arccos \frac{\sqrt{17}-1}{4}+\pi k, t_{2}=\frac{\pi}{4}(2 k+1), t_{3}= \pm \frac{\pi}{6}+\pi k$.