समाधान {l}{cos(2x)=0}{y=sin(2x)-1}{z=y}{a=z}{b=a}{c=b}{text{Solvefor}dtext{where}}{d=c}

$$\left. \begin{array} { l } { \cos(2 x) = 0 }\\ { y = \sin(2 x) - 1 }\\ { z = y }\\ { a = z }\\ { b = a }\\ { c = b }\\ { \text{Solve for } d \text{ where} } \\ { d = c } \end{array} \right.$$

x, y, z, a, b, c, d को लागि हल गर्नुहोस्

$$x\in \cup n_{1},\frac{\pi n_{1}}{2}+\frac{\pi }{4}$$

$$\exists n_{1}\in \mathrm{Z}\text{ : }\left(\exists n_{1}\in \mathrm{Z}\text{ : }\left(\exists n_{1}\in \mathrm{Z}\text{ : }\left(\exists n_{1}\in \mathrm{Z}\text{ : }\left(\exists n_{1}\in \mathrm{Z}\text{ : }\left(\exists n_{2}\in \mathrm{Z}\text{ : }n_{1}=2n_{2}\right)\right)\right)\right)\right)\text{ and }\exists n_{1}\in \mathrm{Z}\text{ : }\left(\exists n_{1}\in \mathrm{Z}\text{ : }\left(\exists n_{1}\in \mathrm{Z}\text{ : }\left(\exists n_{1}\in \mathrm{Z}\text{ : }\left(\exists n_{1}\in \mathrm{Z}\text{ : }\left(\exists n_{1}\in \mathrm{Z}\text{ : }\left(\exists n_{1}\in \mathrm{Z}\text{ : }\left(\exists n_{1}\in \mathrm{Z}\text{ : }\left(\exists n_{1}\in \mathrm{Z}\text{ : }\left(\exists n_{1}\in \mathrm{Z}\text{ : }\left(\exists n_{1}\in \mathrm{Z}\text{ : }\left(\exists n_{1}\in \mathrm{Z}\text{ : }\left(\exists n_{1}\in \mathrm{Z}\text{ : }\left(\exists n_{1}\in \mathrm{Z}\text{ : }\left(\exists n_{1}\in \mathrm{Z}\text{ : }\left(\exists n_{1}\in \mathrm{Z}\text{ : }\left(\exists n_{1}\in \mathrm{Z}\text{ : }\left(\exists n_{1}\in \mathrm{Z}\text{ : }\left(\exists n_{1}\in \mathrm{Z}\text{ : }\left(\exists n_{1}\in \mathrm{Z}\text{ : }c=\left(-1\right)^{n_{1}}-1\text{, }\exists n_{2}\in \mathrm{Z}\text{ : }n_{1}=2n_{2}\right)\right)\text{, }\exists n_{1}\in \mathrm{Z}\text{ : }\left(\exists n_{1}\in \mathrm{Z}\text{ : }\left(\exists n_{2}\in \mathrm{Z}\text{ : }n_{1}=2n_{2}\right)\right)\right)\right)\right)\right)\right)\text{, }\exists n_{1}\in \mathrm{Z}\text{ : }\left(\exists n_{1}\in \mathrm{Z}\text{ : }\left(\exists n_{1}\in \mathrm{Z}\text{ : }\left(\exists n_{1}\in \mathrm{Z}\text{ : }\left(\exists n_{1}\in \mathrm{Z}\text{ : }\left(\exists n_{2}\in \mathrm{Z}\text{ : }n_{1}=2n_{2}\right)\right)\right)\right)\right)\text{ and }\exists n_{1}\in \mathrm{Z}\text{ : }\left(\exists n_{1}\in \mathrm{Z}\text{ : }\left(\exists n_{2}\in \mathrm{Z}\text{ : }n_{1}=2n_{2}\right)\right)\right)\right)\right)\right)\right)\right)\text{, }\exists n_{1}\in \mathrm{Z}\text{ : }\left(\exists n_{1}\in \mathrm{Z}\text{ : }\left(\exists n_{1}\in \mathrm{Z}\text{ : }\left(\exists n_{1}\in \mathrm{Z}\text{ : }\left(\exists n_{1}\in \mathrm{Z}\text{ : }\left(\exists n_{1}\in \mathrm{Z}\text{ : }\left(\exists n_{1}\in \mathrm{Z}\text{ : }\left(\exists n_{1}\in \mathrm{Z}\text{ : }\left(\exists n_{1}\in \mathrm{Z}\text{ : }\left(\exists n_{3}\in \mathrm{Z}\text{ : }n_{1}=2n_{3}\right)\text{, }z=0\text{ and }a=\left(-1\right)^{n_{1}}-1\right)\right)\right)\right)\right)\text{, }\exists n_{1}\in \mathrm{Z}\text{ : }\left(\exists n_{1}\in \mathrm{Z}\text{ : }z=\left(-1\right)^{n_{1}}-1\right)\right)\right)\right)\text{ and }\exists n_{1}\in \mathrm{Z}\text{ : }\left(\exists n_{1}\in \mathrm{Z}\text{ : }\left(\exists n_{1}\in \mathrm{Z}\text{ : }\left(\exists n_{1}\in \mathrm{Z}\text{ : }\left(\exists n_{1}\in \mathrm{Z}\text{ : }\left(\exists n_{2}\in \mathrm{Z}\text{ : }n_{1}=2n_{2}\right)\right)\right)\right)\right)\text{ and }\exists n_{1}\in \mathrm{Z}\text{ : }\left(\exists n_{1}\in \mathrm{Z}\text{ : }\left(\exists n_{2}\in \mathrm{Z}\text{ : }n_{1}=2n_{2}\right)\right)\right)\right)\right)\right)\right)\right)\text{ and }\exists n_{1}\in \mathrm{Z}\text{ : }\left(\exists n_{1}\in \mathrm{Z}\text{ : }\left(\exists n_{1}\in \mathrm{Z}\text{ : }\left(\exists n_{1}\in \mathrm{Z}\text{ : }\left(\exists n_{1}\in \mathrm{Z}\text{ : }\left(\exists n_{1}\in \mathrm{Z}\text{ : }\left(\exists n_{1}\in \mathrm{Z}\text{ : }\left(\exists n_{1}\in \mathrm{Z}\text{ : }\left(\exists n_{1}\in \mathrm{Z}\text{ : }\left(\exists n_{1}\in \mathrm{Z}\text{ : }\left(\exists n_{1}\in \mathrm{Z}\text{ : }\left(\exists n_{1}\in \mathrm{Z}\text{ : }\left(\exists n_{1}\in \mathrm{Z}\text{ : }\left(\exists n_{1}\in \mathrm{Z}\text{ : }\left(\exists n_{2}\in \mathrm{Z}\text{ : }n_{1}=2n_{2}\right)\right)\right)\text{ and }\exists n_{1}\in \mathrm{Z}\text{ : }\left(\exists n_{1}\in \mathrm{Z}\text{ : }\left(\exists n_{2}\in \mathrm{Z}\text{ : }n_{1}=2n_{2}\right)\right)\right)\right)\right)\right)\right)\text{ and }\exists n_{1}\in \mathrm{Z}\text{ : }\left(\exists n_{1}\in \mathrm{Z}\text{ : }\left(\exists n_{2}\in \mathrm{Z}\text{ : }n_{1}=2n_{2}\right)\right)\text{ and }\exists n_{1}\in \mathrm{Z}\text{ : }\left(\exists n_{1}\in \mathrm{Z}\text{ : }\left(\exists n_{1}\in \mathrm{Z}\text{ : }\left(\exists n_{1}\in \mathrm{Z}\text{ : }\left(\exists n_{1}\in \mathrm{Z}\text{ : }\left(\exists n_{2}\in \mathrm{Z}\text{ : }n_{1}=2n_{2}\right)\right)\right)\right)\right)\right)\right)\right)\right)\right)\right)\text{, }n_{1}\in \mathrm{Z}\text{, }y=0\text{, }d=0\text{ and }z=0\text{ and }a=0\text{ and }b=0$$