# Thread: how do you solve this?

1. ## how do you solve this?

6. 5(sin 6t - 4cos 4t) = 0

15. sin2x = cosx

(on the interval [0, 2pi])

Thanks!

2. Originally Posted by finalfantasy
6. 5(sin 6t - 4cos 4t) = 0
we have $\sin 6t - 4 \cos 4t = 0$

use the formulas

$\sin nt = \sin [(n - 1)t + t] = \sin (n - 1)t \cos t + \sin t \cos (n - 1)t$

and

$\cos nt = \cos [(n - 1)t + t] = \cos (n - 1)t \cos t - \sin (n - 1)t \sin t$

to keep breaking down the sines and cosines until you get them to $\sin 2t$ and $\cos 2t$. then use the double angle formulas. then simplify and continue. it will be a lot of work, but it is the kind of work where you can keep following your nose

15. sin2x = cosx

(on the interval [0, 2pi])
use the double angle formulas for sine

$\sin 2x = \cos x$

$\Rightarrow 2 \sin x \cos x = cos x$

$\Rightarrow 2 \sin x \cos x - \cos x = 0$

$\Rightarrow \cos x ( 2 \sin x - 1) = 0$

now continue

EDIT: This is my 53th post!!!!!!!

3. Originally Posted by finalfantasy
6. 5(sin 6t - 4cos 4t) = 0
$sin(6t) - 4sin(4t) = 0$

I'd simplify this a little bit and sub in y = 2x:
$sin(3y) - 4sin(2y) = 0$

Now,
$sin(2y) = 2sin(y)cos(y)$

$sin(3y) = sin(2y + y) = sin(2y)cos(y) + sin(y)cos(2y)$

$= 2sin(y)cos^2(y) + sin(y) - 2sin^3(y)$

$= 2sin(y)(1 - sin^2(y)) + sin(y) - 2sin^3(y)$

$= 3sin(y) - 4sin^3(y)$

So we need to solve
$(3sin(y) - 4sin^3(y)) - 4(2sin(y)cos(y)) = 0$

$-4sin^3(y) + 3sin(y) - 8sin(y)cos(y) = 0$

You can immediately factor out a sin(y), so $y = 0, \pi \implies x = 0, \frac{\p}{2}$ are solutions.

Now we need to solve
$-4sin^2(y) + 3 - 8cos(y) = 0$

Well,
$-4(1 - cos^2(y)) + 3 - 8cos(y) = 0$

$-4 - 4cos^2(y) + 3 - 8cos(y) = 0$

$4cos^2(y) + 8cos(y) +1 = 0$
which is a quadratic in cos(y). Can you take over?

-Dan