Can this trigonometric equation be solved for x_1 and x_2 in terms of \alpha?

[physicsRookie]

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?

 

[cepheid]

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.

 

[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...

 

[HallsofIvy]

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?

 

[physicsRookie]

HallsofIvy, thanks.

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

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.