trigo equation

**cakeboby** · December 18th 2009, 07:18 AM

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

Many thanks!

**mathaddict** · December 18th 2009, 07:29 AM

Originally Posted by cakeboby

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

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 ?

**Grandad** · December 18th 2009, 07:59 AM

Originally Posted by mathaddict

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 $\cos^4x$ .

Grandad

**cakeboby** · December 19th 2009, 10:16 AM

Thanks for all!

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