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
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
$\displaystyle 2\sin(2x) = 3[\cos(2x) + \sin{x} - 1]$
$\displaystyle 4\sin{x}\cos{x} = 3[1 - 2\sin^2{x} + \sin{x} - 1]$
$\displaystyle 4\sin{x}\cos{x} = 3[2\sin^2{x} + \sin{x}]$
$\displaystyle 4\sin{x}\cos{x} = 6\sin^2{x} + 3\sin{x}$
$\displaystyle 0 = 6\sin^2{x} + 3\sin{x} - 4\sin{x}\cos{x}$
$\displaystyle 0 = \sin{x}[6\sin{x} + 3 - 4\cos{x}]$
$\displaystyle \sin{x} = 0$ 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?