text: ## help: trigonometric equation
  
  $$x_1(cos\alpha-1) + x_2sin\alpha = 0$$
  $$x_1sin\alpha + x_2(-cos\alpha-1) = 0$$
  How to solve this equation? Can anyone help me?
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   Mentor 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.
   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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  ## 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 $$cos\alpha= -1$$?
  
  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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