# Help: trigonometric equation

by physicsRookie
Tags: equation, trigonometric
 P: 5 $$x_1(cos\alpha-1) + x_2sin\alpha = 0$$ $$x_1sin\alpha + x_2(-cos\alpha-1) = 0$$ How to solve this equation? Can anyone help me?
 Emeritus Sci Advisor PF Gold P: 5,196 It's a system of equations: 2 equations in 2 unknowns. That means you can solve it. Just solve for one unknown in terms of the other using the first equation, and then subsitute that into the second.
 P: 5 Let me try... Solve equation 1: $$x_2 = \frac{-x_1(cos\alpha - 1)}{sin\alpha}$$ Substitute it to the second: $$x_1sin\alpha + \frac{-x_1(cos\alpha - 1)}{sin\alpha}(-cos\alpha-1) = 0$$ $$x_1sin\alpha + \frac{x_1(cos^2\alpha - 1)}{sin\alpha} = 0$$ $$2x_1sin\alpha = 0$$ What is the solutions for $$2x_1sin\alpha = 0$$? Obviously one is $$x_1=0$$, but if $$sin\alpha = 0$$, then...
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PF Gold
P: 39,682
Help: trigonometric equation

 Quote by physicsRookie Let me try... Solve equation 1: $$x_2 = \frac{-x_1(cos\alpha - 1)}{sin\alpha}$$ Substitute it to the second: $$x_1sin\alpha + \frac{-x_1(cos\alpha - 1)}{sin\alpha}(-cos\alpha-1) = 0$$ $$x_1sin\alpha + \frac{x_1(cos^2\alpha - 1)}{sin\alpha} = 0$$ $$2x_1sin\alpha = 0$$ What is the solutions for $$2x_1sin\alpha = 0$$? Obviously one is $$x_1=0$$, but if $$sin\alpha = 0$$, then...
Well done. The point is, of course, that $$\alpha$$ is a number (not one of the variables) so these can be solved like any pair of equations for x1 and x2.
Notice, by the way, that if $$sin\alpha= 0$$, your first step, dividing by that, would be invalid. You have to look at this case separately.
If $$sin\alpha= 0$$, then $$cos\alpha$$ is either 1 or -1.

What do your equations look like if $$sin\alpha= 0$$ and $$cos\alpha= 1$$?

What do your equations look like if $$sin\alpha= 0$$ and $$cos\alpha= -1$$?
P: 5
HallsofIvy, thanks.
 Quote by HallsofIvy What do your equations look like if $$sin\alpha= 0$$ and $$cos\alpha= 1$$?
$$2x_1=0 and 0=0 => x_1=0, x_2$$ could be any number
 Quote by HallsofIvy What do your equations look like if $$sin\alpha= 0$$ and $$cos\alpha= -1$$?
$$0=0 and -2x_2=0 => x_2=0, x_1$$ could be any number

I just try another solution.
Rewrite the equations:
$$(x_1cos\alpha + x_2sin\alpha) - x_1 = 0$$
$$(x_1sin\alpha - x_2cos\alpha) - x_2 = 0$$

Suppose $$x_1 = cos\frac{\alpha}{2}, x_2 = sin\frac{\alpha}{2}$$, then

$$cos\frac{\alpha}{2}cos\alpha + sin\frac{\alpha}{2}sin\alpha - cos\frac{\alpha}{2} = 0$$

$$cos\frac{\alpha}{2}sin\alpha - sin\frac{\alpha}{2}cos\alpha - sin\frac{\alpha}{2} = 0$$

It works!

I am wondering whether there is some general method to solve $$x_1, x_2$$ depending on $$\alpha$$ or not.

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