# need some help with solving a trig equation

• Nov 16th 2008, 09:14 AM
andyp999
need some help with solving a trig equation
ok, i am supposed to solve

2 sin 2 x = 3(cos 2x + sin x -1) for 0 ≤ x < pi (so in radians)

any help at all? i'm goin to have to move this urgent
• Nov 16th 2008, 12:14 PM
skeeter
Quote:

Originally Posted by andyp999
ok i asked this in the trig forum but havnt had a reply for a few hours and i really need to get this done soon

i am supposed to solve

2 sin 2 x = 3(cos 2x + sin x -1) for 0 ≤ x < pi (so in radians)

confirm the equation ...

$2\sin(2x) = 3[\cos(2x) + \sin{x} - 1]$
• Nov 16th 2008, 12:21 PM
andyp999
sorry for the late reply but yes that is correct
• Nov 16th 2008, 01:01 PM
skeeter
$2\sin(2x) = 3[\cos(2x) + \sin{x} - 1]$

$4\sin{x}\cos{x} = 3[1 - 2\sin^2{x} + \sin{x} - 1]$

$4\sin{x}\cos{x} = 3[2\sin^2{x} + \sin{x}]$

$4\sin{x}\cos{x} = 6\sin^2{x} + 3\sin{x}$

$0 = 6\sin^2{x} + 3\sin{x} - 4\sin{x}\cos{x}$

$0 = \sin{x}[6\sin{x} + 3 - 4\cos{x}]$

$\sin{x} = 0$ at $x = 0$

that was the "easy" solution ... the next is a pain in the @ ...

$6\sin{x} + 3 - 4\cos{x} = 0$

$6\sin{x} = 4\cos{x} - 3$

$36\sin^2{x} = 16\cos^2{x} - 24\cos{x} + 9$

$36(1 - \cos^2{x}) = 16\cos^2{x} - 24\cos{x} + 9$

$36 - 36\cos^2{x} = 16\cos^2{x} - 24\cos{x} + 9$

$0 = 52\cos^2{x} - 24\cos{x} - 27$

$\cos{x} = \frac{6 \pm 3\sqrt{43}}{26}$

one of these two solutions is extraneous ... the solution that works is not pretty, you'll need a calculator to determine x.

$x = \arccos\left(\frac{6 - 3\sqrt{43}}{26}\right)$

which makes me ask the question ... were you allowed to use a calculator in solving this equation in the first place?
• Nov 16th 2008, 01:04 PM
andyp999
yes i was allowed a calculator haha
• Nov 16th 2008, 01:10 PM
skeeter
Quote:

Originally Posted by andyp999
yes i was allowed a calculator haha

not funny.
• Nov 16th 2008, 01:13 PM
andyp999
sorry did u not kno? sorry if i caused you problems