حل y=sin(x+pi/3)cos(pi/6-x)sinpi/4

$$y = \sin ( x + \frac { \pi } { 3 } ) \cos ( \frac { \pi } { 6 } - x ) \sin \frac { \pi } { 4 }$$

برای x حل کنید

$$\left\{\begin{matrix}x=\frac{6\arccos(\sqrt[4]{2}\sqrt{y})+12\pi n_{1}-5\pi }{6}\text{, }n_{1}\in \mathrm{Z}\text{; }x=\frac{-6\arccos(\sqrt[4]{2}\sqrt{y})+12\pi n_{2}+7\pi }{6}\text{, }n_{2}\in \mathrm{Z}\text{, }&y\leq \frac{\sqrt{2}}{2}\text{ and }y\geq 0\text{ and }-\sqrt[4]{2}\sqrt{y}\geq -1\\x=\frac{-6\arccos(\sqrt[4]{2}\sqrt{y})+12\pi n_{3}+\pi }{6}\text{, }n_{3}\in \mathrm{Z}\text{; }x=\frac{6\arccos(\sqrt[4]{2}\sqrt{y})+12\pi n_{4}+\pi }{6}\text{, }n_{4}\in \mathrm{Z}\text{, }&y\leq \frac{\sqrt{2}}{2}\text{ and }y\geq 0\text{ and }\sqrt[4]{2}\sqrt{y}\leq 1\end{matrix}\right.$$

برای y حل کنید

$$y=\frac{\sqrt{2}\left(\cos(\frac{6x-\pi }{6})\right)^{2}}{2}$$