Is x*cos(Ax) + B*sin(Ax) = 0 Solvable for x?

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The equation x*cos(Ax) + B*sin(Ax) = 0 has a trivial solution at x=0, but other solutions exist when x*cos(Ax) = -B*sin(Ax) for x≠0. This form suggests an infinite number of solutions, yet the equation appears intractable for algebraic methods. Simplifying to tan(Ax)/x = -1/B reveals that it lacks an algebraic solution. Therefore, the equation is not solvable using standard algebraic techniques.
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Homework Equations



x*cos(Ax) + B*sin(Ax) = 0

Where A and B are nonzero constants.

The Attempt at a Solution



The only value of x where both terms are zero is x=0. All other solutions (presumably an infinite number of them) occur when:

x*cos(Ax) = -B*sin(Ax), x≠0.

This seems intractable to me, but I don't have the math skills required to recognize an equation that cannot be solved algebraically.
 
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Hi Michael! :smile:
Michael12345 said:
x*cos(Ax) = -B*sin(Ax), x≠0.

This seems intractable to me, but I don't have the math skills required to recognize an equation that cannot be solved algebraically.

You can simplify it as tan(Ax)/x = -1/B …

yes, that has no algebraic solution.
 
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