x^2+y^2(cosx)=8 सोडवा

$$x ^ { 2 } + y ^ { 2 } ( \cos x ) = 8$$

y साठी सोडवा (जटिल उत्तर)

$$y=-\frac{\sqrt{2}i\sqrt{2\left(x^{2}-8\right)}\left(\cos(x)\right)^{-\frac{1}{2}}}{2}$$

$$y=\frac{\sqrt{2}i\sqrt{2\left(x^{2}-8\right)}\left(\cos(x)\right)^{-\frac{1}{2}}}{2}\text{, }\nexists n_{1}\in \mathrm{Z}\text{ : }x=\pi n_{1}+\frac{\pi }{2}$$

y साठी सोडवा

$$y=\sqrt{-\frac{x^{2}-8}{\cos(x)}}$$

$$y=-\sqrt{-\frac{x^{2}-8}{\cos(x)}}\text{, }\left(x>-\frac{3\pi }{2}\text{ and }x\leq -2\sqrt{2}\right)\text{ or }\left(x\geq 2\sqrt{2}\text{ and }x<\frac{3\pi }{2}\right)\text{ or }\left(\exists n_{2}\in \mathrm{Z}\text{ : }\left(x>2\pi n_{2}+\frac{\pi }{2}\text{ and }x<2\pi n_{2}+\frac{3\pi }{2}\right)\text{ and }\nexists n_{1}\in \mathrm{Z}\text{ : }x=\pi n_{1}+\frac{\pi }{2}\right)$$