ok, i am supposed to solve

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

please help me out, i suck at these sort of questions

any help at all? i'm goin to have to move this urgent

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- Nov 16th 2008, 08:14 AMandyp999need 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)

please help me out, i suck at these sort of questions

any help at all? i'm goin to have to move this urgent - Nov 16th 2008, 11:14 AMskeeter
- Nov 16th 2008, 11:21 AMandyp999
sorry for the late reply but yes that is correct

- Nov 16th 2008, 12:01 PMskeeter
$\displaystyle 2\sin(2x) = 3[\cos(2x) + \sin{x} - 1]$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

which makes me ask the question ... were you allowed to use a calculator in solving this equation in the first place? - Nov 16th 2008, 12:04 PMandyp999
yes i was allowed a calculator haha

- Nov 16th 2008, 12:10 PMskeeter
- Nov 16th 2008, 12:13 PMandyp999
sorry did u not kno? sorry if i caused you problems