c^2=1+sinθ+sin^2θ/2! सोडवा

$$c ^ { 2 } = 1 + \sin \theta + \frac { \sin ^ { 2 } \theta } { 2 ! }$$

c साठी सोडवा (जटिल उत्तर)

$$c=-\frac{\sqrt{4\sin(\theta )-\cos(2\theta )+5}}{2}$$

$$c=\frac{\sqrt{4\sin(\theta )-\cos(2\theta )+5}}{2}$$

c साठी सोडवा

$$c=\frac{\sqrt{2\left(\left(\sin(\theta )\right)^{2}+2\sin(\theta )+2\right)}}{2}$$

$$c=-\frac{\sqrt{2\left(\left(\sin(\theta )\right)^{2}+2\sin(\theta )+2\right)}}{2}$$

θ साठी सोडवा

$$\left\{\begin{matrix}\theta =-\arcsin(-\sqrt{2c^{2}-1}+1)+2\pi n_{3}\text{, }n_{3}\in \mathrm{Z}\text{; }\theta =\arcsin(-\sqrt{2c^{2}-1}+1)+2\pi n_{4}+\pi \text{, }n_{4}\in \mathrm{Z}\text{, }&\left(c\geq -\frac{\sqrt{10}}{2}\text{ and }c\leq -\frac{\sqrt{2}}{2}\right)\text{ or }\left(c\geq \frac{\sqrt{2}}{2}\text{ and }c\leq \frac{\sqrt{10}}{2}\right)\\\theta =-\arcsin(\sqrt{2c^{2}-1}+1)+2\pi n_{1}\text{, }n_{1}\in \mathrm{Z}\text{; }\theta =\arcsin(\sqrt{2c^{2}-1}+1)+2\pi n_{2}+\pi \text{, }n_{2}\in \mathrm{Z}\text{, }&|c|=\frac{\sqrt{2}}{2}\end{matrix}\right.$$