Let's solve **Question 49** step-by-step: **49.** \( r = \sin(2\theta) \) and \( r = \cos\theta \) on \( [0, \pi] \) ### **Step 1: Set the equations equal to each other** \[ \sin(2\theta) = \cos\theta \] ### **Step 2: Express \(\sin(2\theta)\) using the double angle identity** \[ \sin(2\theta) = 2 \sin\theta \cos\theta \] So the equation becomes: \[ 2 \sin\theta \cos\theta = \cos\theta \] ### **Step 3: Solve for \(\theta\)** If \(\cos \theta \neq 0\), divide both sides by \(\cos \theta\): \[ 2\sin\theta = 1 \implies \sin\theta = \frac{1}{2} \] So, \[ \theta = \frac{\pi}{6}, \quad \theta = \frac{5\pi}{6} \] (both are in \([0, \pi]\)) If \(\cos\theta = 0\): \[ \cos\theta = 0 \implies \theta = \frac{\pi}{2} \] (\(\theta\) is in \([0, \pi]\)) ### **Step 4: Find \(r\) for each solution** #### For \(\theta = \frac{\pi}{6}\): \[ r = \sin(2\cdot\frac{\pi}{6}) = \sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2} \] or \[ r = \cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2} \] #### For \(\theta = \frac{5\pi}{6}\): \[ r = \sin(2\cdot\frac{5\pi}{6}) = \sin(\frac{5\pi}{3}) = -\frac{\sqrt{3}}{2} \] or \[ r = \cos(\frac{5\pi}{6}) = -\frac{\sqrt{3}}{2} \] #### For \(\theta = \frac{\pi}{2}\): \[ r = \sin(2\cdot\frac{\pi}{2}) = \sin(\pi) = 0 \] or \[ r = \cos(\frac{\pi}{2}) = 0 \] ### **Step 5: Write the intersection points** The points of intersection in polar coordinates \((r, \theta)\) are: \[ \boxed{ \left( \frac{\sqrt{3}}{2}, \frac{\pi}{6} \right),\ \left( -\frac{\sqrt{3}}{2}, \frac{5\pi}{6} \right),\ \left( 0, \frac{\pi}{2} \right) } \] --- **If you want the solution for another question, let me know!**

Certainly! Below are the detailed step-by-step solutions for each of the given exercises involving the intersection points of the polar graphs. --- ### **Question 49:** \[ r = \sin(2\theta) \quad \text{and} \quad r = \cos \theta \quad \text{on} \quad [0, \pi] \] **Step 1:** Set the equations equal to find intersection points: \[ \sin(2\theta) = \cos \theta \] **Step 2:** Use the double angle identity: \[ \sin(2\theta) = 2 \sin \theta \cos \theta \] So, \[ 2 \sin \theta \cos \theta = \cos \theta \] **Step 3:** Solve for \(\theta\): - If \(\cos \theta \neq 0\), divide both sides by \(\cos \theta\): \[ 2 \sin \theta = 1 \implies \sin \theta = \frac{1}{2} \] - Solutions for \(\sin \theta = \frac{1}{2}\) in \([0, \pi]\): \[ \boxed{ \theta = \frac{\pi}{6}, \quad \theta = \frac{5\pi}{6} } \] - If \(\cos \theta = 0\): \[ \cos \theta = 0 \implies \theta = \frac{\pi}{2} \] **Step 4:** Find corresponding \(r\) values: - For \(\theta = \frac{\pi}{6}\): \[ r = \sin(2 \times \frac{\pi}{6}) = \sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2} \] and \[ r = \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2} \] - For \(\theta = \frac{5\pi}{6}\): \[ r = \sin(2 \times \frac{5\pi}{6}) = \sin(\frac{5\pi}{3}) = -\frac{\sqrt{3}}{2} \] and \[ r = \cos \frac{5\pi}{6} = -\frac{\sqrt{3}}{2} \] - For \(\theta = \frac{\pi}{2}\): \[ r = \sin(\pi) = 0, \quad r = \cos \frac{\pi}{2} = 0 \] **Final intersection points:** \[ \boxed{ \left( \frac{\sqrt{3}}{2}, \frac{\pi}{6} \right), \quad \left( -\frac{\sqrt{3}}{2}, \frac{5\pi}{6} \right), \quad (0, \frac{\pi}{2}) } \] --- ### **Question 50:** \[ r = \cos(2\theta) \quad \text{and} \quad r = \cos \theta \quad \text{on} \quad [0, 2\pi] \] **Step 1:** Set equal: \[ \cos(2\theta) = \cos \theta \] **Step 2:** Use the double angle identity: \[ \cos(2\theta) = 2 \cos^2 \theta - 1 \] So, \[ 2 \cos^2 \theta - 1 = \cos \theta \] **Step 3:** Rearrange into quadratic form: \[ 2 \cos^2 \theta - \cos \theta - 1 = 0 \] Let \(x = \cos \theta\), then: \[ 2x^2 - x - 1 = 0 \] Solve using quadratic formula: \[ x = \frac{1 \pm \sqrt{1^2 - 4 \times 2 \times (-1)}}{2 \times 2} = \frac{1 \pm \sqrt{1 + 8}}{4} = \frac{1 \pm \sqrt{9}}{4} \] \[ x = \frac{1 \pm 3}{4} \] - When \(+\): \[ x = \frac{1 + 3}{4} = 1 \] - When \(-\): \[ x = \frac{1 - 3}{4} = -\frac{1}{2} \] **Step 4:** Find \(\theta\): - For \(x = 1\): \[ \cos \theta = 1 \implies \theta = 0 \] (since \(\cos \theta = 1\) at \(\theta=0\)) - For \(x = -\frac{1}{2}\): \[ \cos \theta = -\frac{1}{2} \implies \theta = \frac{2\pi}{3}, \quad \frac{4\pi}{3} \] **Step 5:** Find \(r\) at these \(\theta\): - At \(\theta=0\): \[ r = \cos 0 = 1 \] - At \(\theta = \frac{2\pi}{3}\): \[ r = \cos(2 \times \frac{2\pi}{3}) = \cos \frac{4\pi}{3} = -\frac{1}{2} \] - At \(\theta = \frac{4\pi}{3}\): \[ r = \cos(2 \times \frac{4\pi}{3}) = \cos \frac{8\pi}{3} = \cos \left( 2\pi + \frac{2\pi}{3} \right) = \cos \frac{2\pi}{3} = -\frac{1}{2} \] **Final intersection points:** \[ \boxed{ (1, 0), \quad \left( -\frac{1}{2}, \frac{2\pi}{3} \right), \quad \left( -\frac{1}{2}, \frac{4\pi}{3} \right) } \] --- ### **Question 51:** \[ r=2 \cos \theta \quad \text{and} \quad r=2 \sin \theta \quad \text{on} \quad [0, 2\pi] \] **Step 1:** Set equal: \[ 2 \cos \theta = 2 \sin \theta \implies \cos \theta = \sin \theta \] **Step 2:** Solve: \[ \cos \theta = \sin \theta \implies \tan \theta = 1 \] \[ \boxed{ \theta = \frac{\pi}{4}, \quad \frac{5\pi}{4} } \] **Step 3:** Find \(r\): - At \(\theta = \frac{\pi}{4}\): \[ r = 2 \cos \frac{\pi}{4} = 2 \times \frac{\sqrt{2}}{2} = \sqrt{2} \] or equivalently, \[ r = 2 \sin \frac{\pi}{4} = \sqrt{2} \] - At \(\theta = \frac{5\pi}{4}\): \[ r = 2 \cos \frac{5\pi}{4} = 2 \times \left(-\frac{\sqrt{2}}{2}\right) = -\sqrt{2} \] or \[ r = 2 \sin \frac{5\pi}{4} = -\sqrt{2} \] **Final points:** \[ \boxed{ \left( \sqrt{2}, \frac{\pi}{4} \right), \quad \left( -\sqrt{2}, \frac{5\pi}{4} \right) } \] --- ### **Question 52:** \[ r = 2 \cos \theta \quad \text{and} \quad r = \frac{\sqrt{3}}{3} + 3 \sin \theta \quad \text{on} \quad [0, 2\pi] \] **Step 1:** Set equal: \[ 2 \cos \theta = \frac{\sqrt{3}}{3} + 3 \sin \theta \] **Step 2:** Rearrange: \[ 2 \cos \theta - 3 \sin \theta = \frac{\sqrt{3}}{3} \] **Step 3:** Express as a single sinusoid: - Find \(A\) and \(\phi\): \[ A = \sqrt{(2)^2 + (-3)^2} = \sqrt{4 + 9} = \sqrt{13} \] \[ \cos \phi = \frac{2}{A} = \frac{2}{\sqrt{13}}, \quad \sin \phi = \frac{-3}{A} = -\frac{3}{\sqrt{13}} \] - Rewrite as: \[ A \left( \cos \phi \cos \theta + \sin \phi \sin \theta \right) = \frac{\sqrt{3}}{3} \] \[ \Rightarrow \sqrt{13} \cos (\theta + \phi) = \frac{\sqrt{3}}{3} \] **Step 4:** Solve for \(\theta + \phi\): \[ \cos (\theta + \phi) = \frac{\sqrt{3}}{3 \sqrt{13}} = \frac{\sqrt{3}}{\sqrt{13} \times 3} \] The solutions are: \[ \theta + \phi = \pm \arccos \left( \frac{\sqrt{3}}{3 \sqrt{13}} \right) + 2\pi k \] and \[ \theta = - \phi \pm \arccos \left( \frac{\sqrt{3}}{3 \sqrt{13}} \right) + 2\pi k \] where \[ \phi = \arctan \left( \frac{-3}{2} \right) \] **Note:** Since this involves complicated values, exact solutions are expressed in inverse trigonometric form. --- ### **Question 53:** \[ r = \sin(3\theta) \quad \text{and} \quad r = \cos(3\theta) \quad \text{on} \quad [0, 2\pi] \] **Step 1:** Set equal: \[ \sin(3\theta) = \cos(3\theta) \] **Step 2:** Solve: \[ \tan(3\theta) = 1 \implies 3\theta = \frac{\pi}{4} + n\pi \] for \(n \in \mathbb{Z}\). **Step 3:** Find \(\theta\): \[ \theta = \frac{\pi}{12} + \frac{n \pi}{3} \] for \(n\) such that \(\theta \in [0, 2\pi]\): - For \(n=0\): \[ \theta = \frac{\pi}{12} \] - For \(n=1\): \[ \theta = \frac{\pi}{12} + \frac{\pi}{3} = \frac{\pi}{12} + \frac{4\pi}{12} = \frac{5\pi}{12} \] - For \(n=2\): \[ \theta = \frac{\pi}{12} + \frac{8\pi}{12} = \frac{9\pi}{12} = \frac{3\pi}{4} \] - For \(n=3\): \[ \theta = \frac{\pi}{12} + \frac{12\pi}{12} = \frac{\pi}{12} + \pi = \frac{\pi}{12} + \frac{12\pi}{12} = \frac{13\pi}{12} \] - For \(n=4\): \[ \theta = \frac{\pi}{12} + \frac{16\pi}{12} = \frac{\pi}{12} + \frac{4\pi}{3} = \frac{\pi}{12} + \frac{16\pi}{12} = \frac{17\pi}{12} \] All these are within \([0, 2\pi]\). **Final solutions:** \[ \boxed{ \theta = \frac{\pi}{12}, \quad \frac{5\pi}{12}, \quad \frac{3\pi}{4}, \quad \frac{13\pi}{12}, \quad \frac{17\pi}{12} } \] --- ### **Question 54:** \[ r = 3 \cos \theta \quad \text{and} \quad r = 1 + \cos \theta \quad \text{on} \quad [0, 2\pi] \] **Step 1:** Set equal: \[ 3 \cos \theta = 1 + \cos \theta \] **Step 2:** Solve: \[ 3 \cos \theta - \cos \theta = 1 \implies 2 \cos \theta = 1 \implies \cos \theta = \frac{1}{2} \] **Step 3:** Find \(\theta\): \[ \theta = \pm \frac{\pi}{3} \] Within \([0, 2\pi]\), solutions are: \[ \boxed{ \theta = \frac{\pi}{3}, \quad \frac{5\

Certainly! Continuing from where I left off: ### **Question 54 (continued):** We found: \[ \cos \theta = \frac{1}{2} \implies \theta = \frac{\pi}{3}, \quad \frac{5\pi}{3} \] ### **Step 4:** Find the corresponding \(r\) values: - At \(\theta = \frac{\pi}{3}\): \[ r = 3 \cos \frac{\pi}{3} = 3 \times \frac{1}{2} = \frac{3}{2} \] or \[ r = 1 + \cos \frac{\pi}{3} = 1 + \frac{1}{2} = \frac{3}{2} \] - At \(\theta = \frac{5\pi}{3}\): \[ r = 3 \cos \frac{5\pi}{3} = 3 \times \frac{1}{2} = \frac{3}{2} \] or \[ r = 1 + \cos \frac{5\pi}{3} = 1 + \frac{1}{2} = \frac{3}{2} \] ### **Final intersection points:** \[ \boxed{ \left( \frac{3}{2}, \frac{\pi}{3} \right), \quad \left( \frac{3}{2}, \frac{5\pi}{3} \right) } \] --- ### **Question 55:** \[ r = 1 \quad \text{and} \quad r = 2 \sin(2\theta) \quad \text{on} \quad [0, 2\pi] \] **Step 1:** Set equal: \[ 1 = 2 \sin(2\theta) \] \[ \Rightarrow \sin(2\theta) = \frac{1}{2} \] **Step 2:** Solve for \(2\theta\): \[ 2 \theta = \arcsin \frac{1}{2} = \frac{\pi}{6}, \quad \frac{5\pi}{6} \] Within \([0, 2\pi]\), \(\theta\) is: \[ \theta = \frac{\pi}{12}, \quad \frac{5\pi}{12} \] from the first solution, and also: \[ 2 \theta = \pi - \frac{\pi}{6} = \frac{5\pi}{6} \] which gives \[ \theta = \frac{5\pi}{12} \] (already found), and \[ 2 \theta = \pi - \frac{5\pi}{6} = \frac{\pi}{6} \] which again gives \[ \theta = \frac{\pi}{12} \] Similarly, for the other principal solutions, the set remains: \[ \boxed{ \theta = \frac{\pi}{12}, \quad \frac{5\pi}{12} } \] **Note:** Since \(\sin(2\theta) = \frac{1}{2}\) has two solutions per period, the points are these two values. --- ### **Question 56:** \[ r = \sin \theta \quad \text{and} \quad r = \frac{1}{3} + 2 \sin \theta \quad \text{on} \quad [0, 2\pi] \] **Step 1:** Set equal: \[ \sin \theta = \frac{1}{3} + 2 \sin \theta \] \[ \Rightarrow \sin \theta - 2 \sin \theta = \frac{1}{3} \] \[ - \sin \theta = \frac{1}{3} \] \[ \sin \theta = - \frac{1}{3} \] **Step 2:** Solve for \(\theta\): \[ \theta = \arcsin \left( - \frac{1}{3} \right) \] Since \(\sin \theta = - \frac{1}{3}\), the solutions in \([0, 2\pi]\): \[ \theta = \arcsin \left( - \frac{1}{3} \right), \quad \pi - \arcsin \left( - \frac{1}{3} \right) \] But because \(\arcsin (-1/3)\) is negative, in \([0, 2\pi]\): \[ \theta_1 = 2\pi + \arcsin \left( - \frac{1}{3} \right) \] which is approximately: \[ \theta_1 \approx 2\pi - 0.3398 \approx 6.283 - 0.3398 = 5.9432 \] and \[ \theta_2 = \pi - \arcsin \left( - \frac{1}{3} \right) \approx \pi + 0.3398 \approx 3.4814 \] ### **Final solutions:** \[ \boxed{ \theta \approx 3.4814, \quad 5.9432 } \] **Corresponding \(r\) values:** - For either value, \(r = \sin \theta\): \[ r \approx - \frac{1}{3} \] --- ## **Summary of solutions:** | Question | Intersection points (in \(\theta, r\)) | Remarks | |------------|------------------------------------------|---------| | 49 | \(\left( \frac{\sqrt{3}}{2}, \frac{\pi}{6} \right)\), \(\left( -\frac{\sqrt{3}}{2}, \frac{5\pi}{6} \right)\), \((0, \frac{\pi}{2})\) | Solved via setting equal and identities | | 50 | \(\left( 1, 0 \right), \left( -\frac{1}{2}, \frac{2\pi}{3} \right), \left( -\frac{1}{2}, \frac{4\pi}{3} \right)\) | Quadratic in \(\cos \theta\) | | 51 | \(\left( \sqrt{2}, \frac{\pi}{4} \right), \left( -\sqrt{2}, \frac{5\pi}{4} \right)\) | \(\tan \theta = 1\) solutions | | 52 | \(\theta = - \phi \pm \arccos(\text{value})\) | Requires inverse trig functions | | 53 | \(\theta = \frac{\pi}{12} + \frac{n\pi}{3}\), \(n=0,\dots,4\) | \(\tan 3\theta = 1\) solutions | | 54 | \(\theta = \frac{\pi}{3}, \frac{5\pi}{3}\) | \(\cos \theta = \frac{1}{2}\) | | 55 | \(\theta = \frac{\pi}{12}, \frac{5\pi}{12}\) | \(\sin 2\theta = \frac{1}{2}\) | | 56 | \(\theta \approx 3.48, 5.94\) | \(\sin \theta = -\frac{1}{3}\) | --- If you need further clarification or solutions in rectangular coordinates, just ask!

Certainly! Here are the final answers for each exercise: **49:** \(\boxed{ \left( \frac{\sqrt{3}}{2}, \frac{\pi}{6} \right),\ \left( -\frac{\sqrt{3}}{2}, \frac{5\pi}{6} \right),\ (0, \frac{\pi}{2}) }\) **50:** \(\boxed{ (1, 0),\ \left( -\frac{1}{2}, \frac{2\pi}{3} \right),\ \left( -\frac{1}{2}, \frac{4\pi}{3} \right) }\) **51:** \(\boxed{ \left( \sqrt{2}, \frac{\pi}{4} \right),\ \left( -\sqrt{2}, \frac{5\pi}{4} \right) }\) **52:** \[ \theta = - \arctan\left(\frac{3}{2}\right) \pm \arccos\left(\frac{\sqrt{13}}{13}\right) \] **53:** \(\boxed{ \theta = \frac{\pi}{12} + \frac{n\pi}{3},\ n=0,1,2,3,4 }\) **54:** \(\boxed{ \theta = \frac{\pi}{3},\ \frac{5\pi}{3} }\) **55:** \(\boxed{ \theta \approx 3.48,\ 5.94 }\) **56:** \(\boxed{ \theta \approx 3.48,\ 5.94 }\) Let me know if you'd like any further assistance!