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I have this expression $$(-Ax\sin(xt) + Bx\cos(x t))\times|-Ax\sin(x t) + Bx\cos(x t)| = 0.$$

I try to recast it in terms of a simple function of $\cos(x t)$ and $\sin(x t)$. The absolute value blocks me and I don't know what to do with it.

Thank you for your advice!

sobat
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  • You have something of the form $f(x,t) |f(x,t)| = 0$. The LHS is 0 whenever $f = 0$, so why not just solve $f(x,t) = 0$. In your case, that means $-Ax \sin (xt) + Bx \cos(xt) = 0$. – Sam Nov 04 '22 at 20:53
  • $y\cdot|y|=\text{sgn}(y)y^2$ – John Wayland Bales Nov 04 '22 at 21:08
  • This equation is part of a bigger system and I have to simplify/recast it to solve it so solving $f(x,t)=0$ it is not really possible... Using $sgn(y)y^2$ gives me a condition on $t$ and I want to get rid of this parameter. – sobat Nov 04 '22 at 21:59

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