다음을 풀어보세요: 3x=asinx+(4-2|a|)sin^2x

$$3 x = a \sin x + ( 4 - 2 | a | ) \sin ^ { 2 } x$$

a에 대한 해

$$\left\{\begin{matrix}a=\frac{-4\left(\sin(x)\right)^{2}+3x}{\sin(x)\left(-2\sin(x)+1\right)}\text{, }&\nexists n_{1}\in \mathrm{Z}\text{ : }x=\pi n_{1}\text{ and }\nexists n_{2}\in \mathrm{Z}\text{ : }\left(x=\frac{\pi \left(12n_{2}+5\right)}{6}\text{ or }x=\frac{\pi \left(12n_{2}+1\right)}{6}\right)\text{ and }\frac{-4\left(\sin(x)\right)^{2}+3x}{\sin(x)\left(-2\sin(x)+1\right)}\geq 0\\a\geq 0\text{, }&\left(-4\left(\sin(x)\right)^{2}+3x=0\text{ and }\exists n_{2}\in \mathrm{Z}\text{ : }\left(x=\frac{\pi \left(12n_{2}+5\right)}{6}\text{ or }x=\frac{\pi \left(12n_{2}+1\right)}{6}\right)\right)\text{ or }\left(-4\left(\sin(x)\right)^{2}+3x=0\text{ and }\exists n_{3}\in \mathrm{Z}\text{ : }x=\pi n_{3}\right)\\a=\frac{-4\left(\sin(x)\right)^{2}+3x}{\sin(x)\left(2\sin(x)+1\right)}\text{, }&\nexists n_{1}\in \mathrm{Z}\text{ : }x=\pi n_{1}\text{ and }\nexists n_{4}\in \mathrm{Z}\text{ : }\left(x=\frac{\pi \left(12n_{4}+11\right)}{6}\text{ or }x=\frac{\pi \left(12n_{4}+7\right)}{6}\right)\text{ and }\frac{-4\left(\sin(x)\right)^{2}+3x}{\sin(x)\left(2\sin(x)+1\right)}\leq 0\\a\leq 0\text{, }&\left(-4\left(\sin(x)\right)^{2}+3x=0\text{ and }\exists n_{4}\in \mathrm{Z}\text{ : }\left(x=\frac{\pi \left(12n_{4}+11\right)}{6}\text{ or }x=\frac{\pi \left(12n_{4}+7\right)}{6}\right)\right)\text{ or }\left(-4\left(\sin(x)\right)^{2}+3x=0\text{ and }\exists n_{5}\in \mathrm{Z}\text{ : }x=\pi n_{5}\right)\end{matrix}\right.$$