Just as the title describes. Is it true or false? I know it is a line in 2D, but circle in 3D.
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$r=3\cos\theta\implies\sqrt{x^2+y^2}=\frac{3x}{\sqrt{x^2+y^2}}$, or $x^2-3x+y^2=0$ where $(x,y)$ is the cartesian coordinates of the point $(r,\theta)$. $x^2-3x+y^2=0$ is not the equation of a straight line. In fact it is the equation of a circle centered at $(3/2,0)$ with radius $3/2$. $$\left(x-\frac32\right)^2+y^2=\left(\frac32\right)^2$$
QED
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$\cos\theta=\frac{x}{\sqrt{x^2+y^2}}$. The factor $3$ comes from $3\cos\theta$. – QED Oct 11 '20 at 21:13
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I forgot the denominator. Thanks! – Bernard Oct 11 '20 at 21:14