# How can I approximately solve trigonometric equations simultaneously?

• Feb 11th 2011, 05:05 PM
Marckutt
How can I approximately solve trigonometric equations simultaneously?
if I have p/k=sin(3x)/sin(2x), how could I solve for x in terms of p/k? I understand that these are not linear functions so they cannot be solved simultaneously but is it possible to approximately solve this problem using taylor expansions? If there is another way I could solve this, I would like to hear what you think. Thank you
• Feb 11th 2011, 06:55 PM
Prove It
You need to use Trigonometric Identities...

$\displaystyle \sin{3x} \equiv \sin{x}(3 - 4\sin^2{x}), \sin{2x} \equiv 2\sin{x}\cos{x}$ and $\displaystyle \sin^2{x} + \cos^2{x} = 1$.

So if $\displaystyle \frac{\sin{3x}}{\sin{2x}} = \frac{p}{k}$

$\displaystyle \frac{\sin{x}(3-4\sin^2{x})}{2\sin{x}\cos{x}} = \frac{p}{k}$

$\displaystyle \frac{3 - 4\sin^2{x}}{2\cos{x}} = \frac{p}{k}$

$\displaystyle \frac{3 - 4(1 - \cos^2{x})}{\cos{x}} = \frac{p}{k}$

$\displaystyle \frac{3 - 4 + 4\cos^2{x}}{\cos{x}} = \frac{p}{k}$

$\displaystyle \frac{4\cos^2{x} - 1}{\cos{x}} = \frac{p}{k}$

$\displaystyle 4\cos^2{x} - 1 = \frac{p}{k}\cos{x}$

$\displaystyle 4\cos^2{x} - \frac{p}{k}\cos{x} - 1 = 0$.

Now let $\displaystyle X = \cos{x}$ and the equation becomes

$\displaystyle 4X^2 - \frac{p}{k}X - 1 = 0$,

which is a quadratic equation that you can solve for $\displaystyle X$, the solution of which you can use to solve for $\displaystyle x$.
• Feb 11th 2011, 07:08 PM
Marckutt
thank you very much