3. Solve the equation $\sin ^{3} x+6 \cos ^{3} x+\frac{1}{\sqrt{2}} \sin 2 x \sin \left(x+\frac{\pi}{4}\right)=0$.
Solution. Transform the equation

$$
\begin{aligned}
& \sin ^{3} x+\sin x \cos x(\sin x+\cos x)+6 \cos ^{3} x=0 \\
& \sin ^{3} x+\sin ^{2} x \cos x+\sin x \cos ^{2} x+6 \cos ^{3} x=0
\end{aligned}
$$

If $\cos x=0$, then substituting into the equation, we get $\sin x=0$, which is impossible. Divide by $\cos ^{3} x$ :

$$
\operatorname{tg}^{3} x+\operatorname{tg}^{2} x+\operatorname{tg} x+6=0
$$

Make the substitution $y=\operatorname{tg} x$ :

$$
\begin{gathered}
y^{3}+y^{2}+y+6=0 \\
(y+2)\left(y^{2}-y+3\right)=0 \\
y=-2, \operatorname{tg} x=-2, x=-\operatorname{arctg} 2+\pi n, n \in Z
\end{gathered}
$$

Answer: $x=-\operatorname{arctg} 2+\pi n, n \in Z$.