Originally Posted by
mathaddict HI
$\displaystyle \frac{6\sin^2 x}{\cos^2 x}-4\sin^2 x-1=0$
$\displaystyle 6\sin^2 x-4\sin^2 x\cos^2 x-\cos^2 x=0$
$\displaystyle 6(1-\cos^2 x)-4(1-\cos^2 x)(\cos^2 x)-\cos^2 x=0$
$\displaystyle 6-6\cos^2 x-4\cos^2 x\color{red}-\cos^4 x\color{black}-\cos^2 x=0$
Can you continue ?