# Math Help - need some help with solving a trig equation

1. ## 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

2. 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 ...

$2\sin(2x) = 3[\cos(2x) + \sin{x} - 1]$

3. sorry for the late reply but yes that is correct

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

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

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

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

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

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

$\sin{x} = 0$ 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?

5. yes i was allowed a calculator haha

6. Originally Posted by andyp999
yes i was allowed a calculator haha
not funny.

7. sorry did u not kno? sorry if i caused you problems