# Solving 13sinx-84sinx = 17

• Feb 24th 2011, 02:19 PM
Resilient
Solving 13sinx-84cosx = 17
Find all $x$ such that $13\sin{x}-84\cos{x} = 17$.

EDIT: the (main) title is wrong, it should be 13sinx-84cosx = 17.
• Feb 24th 2011, 04:26 PM
skeeter
Quote:

Originally Posted by Resilient
Find all $x$ such that $13\sin{x}-84\cos{x} = 17$.

EDIT: the (main) title is wrong, it should be 13sinx-84cosx = 17.

$A\sin{x} - B\cos{x} = C$

$A\sin{x} - B\cos{x} = R\sin(x - \alpha)$

$R = \sqrt{A^2+B^2}$

$\alpha = \arctan\left(\dfrac{B}{A}\right) = \arctan\left(\dfrac{84}{13}\right)$

so, assuming $0 \le x < 2\pi$ ...

$13\sin{x} - 84\cos{x} = 85\sin(x - \alpha)$

$85\sin(x - \alpha) = 17
$

$\sin(x - \alpha) = \dfrac{1}{5}$

$x - \alpha = \arcsin\left(\dfrac{1}{5}\right)$

$x - \alpha = \pi - \arcsin\left(\dfrac{1}{5}\right)$
• Feb 24th 2011, 05:14 PM
Resilient
Quote:

Originally Posted by skeeter
...

Thanks. Can we find $\tan^{-1}\frac{84}{13}$ and $\sin^{-1}\frac{1}{5}$ in terms of $\pi$?
• Feb 25th 2011, 12:35 PM
skeeter
Quote:

Originally Posted by Resilient
Thanks. Can we find $\tan^{-1}\frac{84}{13}$ and $\sin^{-1}\frac{1}{5}$ in terms of $\pi$?

would be really difficult ... use the calculator approximation.
• Feb 25th 2011, 06:04 PM
skoker
@Resilient

I had one just like this and I found a way to get the correct solution with pencil but... your's is a bit to nasty to get a clean radian solution as skeeter says. you can check my thread how I did almost the same problem.

http://www.mathhelpforum.com/math-he...on-172605.html

I got yours into quadratic form in sin function but the coefficients have no common factors other then 1

$13sin\theta-84cos\theta=17$

$(-84\sqrt{1-sin^2\theta})^2=(17-13sin\theta)^2$

$7225sin^2\theta-442sin\theta-6767=0$

$(7225:5^2,17^2)(442:2,13,17)(6767:67,101)$