11.3. For what real values of $\boldsymbol{\alpha}$ does the equation $\sin 3 x=\boldsymbol{\alpha} \sin x+(4-2|\boldsymbol{\alpha}|) \sin ^{2} x$ have the same set of real solutions as the equation $\sin 3 x+\cos 2 x=1+2 \sin x \cos 2 x$?
## Solution.

Using the identities $\sin 3 x=3 \sin x-4 \sin ^{3} x$ and $\cos 2 x=1-2 \sin ^{2} x$, the second equation is equivalent to $\sin x-2 \sin ^{2} x=0$, that is, $\sin x=0$ or $\sin x=\frac{1}{2}$.

Using the identity $\sin 3 x=3 \sin x-4 \sin ^{3} x$, the first equation is equivalent to the equation $3 \sin x-4 \sin ^{3} x=\boldsymbol{\alpha} \sin x+(4-2|\boldsymbol{\alpha}|) \sin ^{2} x$. It is observed that $\sin x=0$ satisfies this equation. For $\sin x \neq 0$, the given equation takes the form $3-4 \sin ^{2} x=\boldsymbol{\alpha}+(4-2|\boldsymbol{\alpha}|) \sin x$. By imposing that $\sin x=\frac{1}{2}$ satisfies this equality, we get $\boldsymbol{\alpha}-|\boldsymbol{\alpha}|=0$, equivalent to $\boldsymbol{\alpha} \geq 0$.

For $\boldsymbol{\alpha} \geq 0$ (and $\sin x \neq 0$), the given equation takes the form $3-4 \sin ^{2} x=\boldsymbol{\alpha}+(4-2 \boldsymbol{\alpha}) \sin x$, equivalent to the equation $4 \sin ^{2} x+(4-2 \boldsymbol{\alpha}) \sin x+(\boldsymbol{\alpha}-3)=0$. Knowing that $\sin x=\frac{1}{2}$ satisfies it, we immediately obtain the equivalent form $2\left(\sin x-\frac{1}{2}\right)(2 \sin x-(\boldsymbol{\alpha}-3))=0$.

Since it is required that the sets of real solutions of the two initial equations coincide, the equation $2 \sin x-(\alpha-3)=0$ must not have other solutions, except those that satisfy the equalities $\sin x=0$ or $\sin x=\frac{1}{2}$. Since the last equation is equivalent to $\sin x=\frac{\boldsymbol{\alpha}-3}{2}$, we must have $\frac{\boldsymbol{\alpha}-3}{2}=0, \frac{\boldsymbol{\alpha}-3}{2}=\frac{1}{2}$ or $\left|\frac{\boldsymbol{\alpha}-3}{2}\right|>1$, equivalent to $\boldsymbol{\alpha}=3, \boldsymbol{\alpha}=4,|\boldsymbol{\alpha}-3|>2$. Since $\alpha \geq 0$, we have $\boldsymbol{\alpha} \in[0,1) \cup\{3,4\} \cup(5,+\infty)$.