1. ## trigo equation

Would anyone please tell me how to do this?
$
6tan^2 x - 4 sin^2 x =1$

Many thanks!

2. Originally Posted by cakeboby
Would anyone please tell me how to do this?
$
6tan^2 x - 4 sin^2 x =1$

Many thanks!
HI

$\frac{6\sin^2 x}{\cos^2 x}-4\sin^2 x-1=0$

$6\sin^2 x-4\sin^2 x\cos^2 x-\cos^2 x=0$

$6(1-\cos^2 x)-4(1-\cos^2 x)(\cos^2 x)-\cos^2 x=0$

$6-6\cos^2 x-4\cos^2 x+4\cos^4 x-\cos^2 x=0$

Can you continue ?

HI

$\frac{6\sin^2 x}{\cos^2 x}-4\sin^2 x-1=0$

$6\sin^2 x-4\sin^2 x\cos^2 x-\cos^2 x=0$

$6(1-\cos^2 x)-4(1-\cos^2 x)(\cos^2 x)-\cos^2 x=0$

$6-6\cos^2 x-4\cos^2 x\color{red}-\cos^4 x\color{black}-\cos^2 x=0$

Can you continue ?
I think you need to check both the sign and the coefficient of the term in $\cos^4x$.