• # question_answer Number of ordered pairs $(x,y)$ satisfying the equation $4{{y}^{2}}+2{{\cos }^{2}}x=4y-{{\sin }^{2}}x$, where $x,y\in [0,2\pi ]$, is A)  1                      B)  2 C)  3                                 D)  4

Given, $4{{y}^{2}}+2{{\cos }^{2}}x=4y-{{\sin }^{2}}x$ $\Rightarrow \,4{{y}^{2}}-4y+1+{{\cos }^{2}}x=0$ $\Rightarrow \,{{(2y-1)}^{2}}+{{\cos }^{2}}x=0$ $\therefore \,\,y=\frac{1}{2}\,\,and\,\,x=0\Rightarrow \,x=\frac{\pi }{2},\,\frac{3\pi }{2}$ So, two ordered pairs are possible i.e., $\left( \frac{\pi }{2},\,\frac{1}{2} \right)$ and $\left( \frac{3\pi }{2},\,\frac{1}{2} \right)$