11.2. Solve the equation: $\sin ^{2} x+\sin ^{2} 2 x+\sin ^{2} 3 x=2$.
Answer. $x_{1}=\frac{\Pi}{4}+\frac{\Pi n}{2}, x_{2}=\frac{\Pi}{2}+\Pi n, x_{1}= \pm \frac{\Pi}{6}+\Pi n$.

Solution. We will use the formula $\sin ^{2} t=\frac{1-\cos 2 t}{2}$, and transform the equation to $\cos 6 x+\cos 4 x+\cos 2 x=-1$. We will add the first and third cosines in it: $2 \cos 2 x \operatorname{Cos} 4 x+\cos 4 x=-1$ and make the substitution $\cos 2 x=t$. We get the equation $2 t^{3}+t^{2}-t=0$, from which $t_{1}=0, t_{2}=-1, t_{3}=\frac{-1}{2}$. From here we find three series of solutions: $x_{1}=\frac{\Pi}{4}+\frac{\Pi n}{2}$, $x_{2}=\frac{\Pi}{2}+\Pi n, x_{1}= \pm \frac{\Pi}{6}+\Pi n$. No selection of roots is needed here.

Grading Criteria. Losing one series of solutions, or finding them incorrectly: minus 3 points. Losing two series, or finding them incorrectly: minus 5 points.