Question

The number of points inside or on the circle $x^2+y^2=4$ satisfying $ta{n}^{2}x+co{t}^{4}x+1=3si{n}^{2}y$ is

• A. One
• B. Two
• C. Four
• D. Infinite

Answer 1

The correct option is C Four

$ta{n}^{4}x+co{t}^{4}x+1=(ta{n}^{2}x-co{t}^{2}x{)}^2+3\ge 3$
The only possible values are $ta{n}^{2}x=co{t}^{2}x$
Hence, $tanx=1$ or $tanx=-1$ and $siny=1$
Hence, $x$ can assume the values - $\frac{-\pi }{4},\frac{\pi }{4}$
$y$ can only assume the values - $\frac{-\pi }{2},\frac{\pi }{2}$
Hence there are $4$ points .