{l}{tan(θ)=5/2}{a=θ}{b=a}{text{Solvefor}ctext{where}}{c=b} ਨੂੰ ਹੱਲ ਕਰੋ

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

θ, a, b, c ਲਈ ਹਲ ਕਰੋ (ਜਟਿਲ ਹੱਲ)

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

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

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

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

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

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

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

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

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

θ, a, b, c ਲਈ ਹਲ ਕਰੋ

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

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