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Prompt: how many times does $r = 2 + sin(\theta)$ pass through x = 1 in cartesian coordinates?

To determine how many times the polar equation $r = 2 + \sin(\theta)$ passes through $x = 1$ in Cartesian coordinates, we need to express the equation in terms of $x$ and $y$.



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The relationship between polar and Cartesian coordinates is given by $x = r\cos(\theta)$ and $y = r\sin(\theta)$.



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Substitute $r$ from the given polar equation $r = 2 + \sin(\theta)$ into the equation for $x$, yielding $x = (2 + \sin(\theta))\cos(\theta)$.



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Since we want to find the instances where $x = 1$, we can set $(2 + \sin(\theta))\cos(\theta) = 1$.



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Expanding and rearranging the equation results in $2\cos(\theta) + \sin(\theta)\cos(\theta) = 1$.



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We can use the double-angle formula for $\sin(2\theta) = 2\sin(\theta)\cos(\theta)$, but here we only have $\sin(\theta)\cos(\theta)$, so we'll leave it as is.



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However, recall the trigonometric identity for $\sin(\theta)\cos(\theta) = \frac{1}{2}\sin(2\theta)$ and rewrite the equation as $2\cos(\theta) + \frac{1}{2}\sin(2\theta) = 1$.



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Unfortunately, solving this equation analytically is complex and may not yield an easily interpretable result. We'll consider another approach.



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Return to the original equation $x = (2 + \sin(\theta))\cos(\theta)$ and recall that $\cos(\theta)$ can range from $-1$ to $1$.



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Since we are interested in $x = 1$, the solutions would be the intersection points of the curve with the line $x = 1$.



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Considering the nature of the polar equation and possible graph shapes, there can be multiple intersections.



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Upon closer inspection and recalling the range of $\sin(\theta)$, it becomes apparent that the graph of $r = 2 + \sin(\theta)$ will intersect the line $x = 1$ twice.



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This is because the curve, as $\theta$ varies, will cross the line $x = 1$ twice due to its shape and periodicity of the trigonometric function.



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Hence, we can deduce that the polar equation $r = 2 + \sin(\theta)$ passes through $x = 1$ twice.



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