# Solve a trigonometric equation

• March 8th 2009, 09:41 AM
antrax1988
Solve a trigonometric equation
34*Sin^2[x]+Sin[2*x]-8*Cos^2[x]==4

Thx
• March 8th 2009, 09:56 AM
red_dog
$34\sin^2x+2\sin x\cos x-8\cos^2x=4(\sin^2x+\cos^2x)\Rightarrow$

$\Rightarrow 30\sin^2x+2\sin x\cos x-12\cos^2x=0| \ :2\cos^2x$

$15\tan^2x+\tan x-6=0$

• March 8th 2009, 09:56 AM
skeeter
Quote:

Originally Posted by antrax1988
34*Sin^2[x]+Sin[2*x]-8*Cos^2[x]==4

Thx

$34\sin^2{x} + \sin(2x) - 8\cos^2{x} = 4(\sin^2{x} + \cos^2{x})
$

$30\sin^2{x} + 2\sin{x}\cos{x} - 12\cos^2{x} = 0$

$15\sin^2{x} + \sin{x}\cos{x} - 6\cos^2{x} = 0$

$(5\sin{x} - 3\cos{x})(3\sin{x} + 2\sin{x}) = 0$

I leave the rest for you ... set each factor equal to zero and solve.
• March 8th 2009, 10:22 AM
antrax1988
Thx guys for the help