{l}{tan(θ)=5/2}{a=θ}{b=a}{c=b}{text{Solvefor}dtext{where}}{d=c} সমাধান কৰক

$$\left. \begin{array} { l } { \tan(\theta) = \frac{5}{2} }\\ { a = \theta }\\ { b = a }\\ { c = b }\\ { \text{Solve for } d \text{ where} } \\ { d = c } \end{array} \right.$$

θ, a, b, c, d-ৰ বাবে সমাধান কৰক (জটিল সমাধান)

$$\theta =\pi n_{1}+\arctan(\frac{5}{2})$$

$$n_{1}\in \mathrm{Z}$$

$$a=\pi n_{1}+\arctan(\frac{5}{2})$$

$$n_{1}\in \mathrm{Z}$$

$$b\in \cup n_{1},\pi n_{1}+\arctan(\frac{5}{2})$$

$$n_{1}\in \mathrm{Z}$$

$$c\in \cup n_{1},\cup n_{1},\cup n_{1},\cup n_{1},\pi n_{1}+\arctan(\frac{5}{2})$$

$$b=\pi n_{1}+\arctan(\frac{5}{2})$$

$$n_{1}\in \mathrm{Z}$$

$$d\in \cup n_{1},\cup n_{1},\cup n_{1},\cup n_{1},\cup n_{1},\cup n_{1},\cup n_{1},\cup n_{1},\pi n_{1}+\arctan(\frac{5}{2})$$

$$c=\pi n_{1}+\arctan(\frac{5}{2})\text{ and }b=\pi n_{1}+\arctan(\frac{5}{2})$$

$$\exists n_{1}\in \mathrm{Z}\text{ : }\left(\exists n_{1}\in \mathrm{Z}\text{ : }b=\pi n_{1}+\arctan(\frac{5}{2})\right)$$

$$n_{1}\in \mathrm{Z}$$

θ, a, b, c, d-ৰ বাবে সমাধান কৰক

$$\theta =2\pi n_{1}+\arcsin(\frac{5\sqrt{29}}{29})+\pi \text{, }n_{1}\in \mathrm{Z}\text{, }a=2\pi n_{1}+\arcsin(\frac{5\sqrt{29}}{29})+\pi \text{, }n_{1}\in \mathrm{Z}\text{, }b=2\pi n_{1}+\arcsin(\frac{5\sqrt{29}}{29})+\pi \text{, }n_{1}\in \mathrm{Z}\text{, }c=2\pi n_{1}+\arcsin(\frac{5\sqrt{29}}{29})+\pi \text{, }n_{1}\in \mathrm{Z}\text{, }d=2\pi n_{1}+\arcsin(\frac{5\sqrt{29}}{29})+\pi \text{, }n_{1}\in \mathrm{Z}$$

$$\theta =2\pi n_{2}+\arcsin(\frac{5\sqrt{29}}{29})\text{, }n_{2}\in \mathrm{Z}\text{, }a=2\pi n_{2}+\arcsin(\frac{5\sqrt{29}}{29})\text{, }n_{2}\in \mathrm{Z}\text{, }b=2\pi n_{2}+\arcsin(\frac{5\sqrt{29}}{29})\text{, }n_{2}\in \mathrm{Z}\text{, }c=2\pi n_{2}+\arcsin(\frac{5\sqrt{29}}{29})\text{, }n_{2}\in \mathrm{Z}\text{, }d=2\pi n_{2}+\arcsin(\frac{5\sqrt{29}}{29})\text{, }n_{2}\in \mathrm{Z}$$