91Ó°ÊÓ

Given function: \[ x = 2 \sin \left(n t + \frac{\pi}{3}\right) \] Use the angle addition formula for sine: \[ \sin(A + B) = \sin A \cos B + \cos A \sin B \] Substitute \( A = n t \) and \( B = \frac{\pi}{3} \): \[ x = 2 \left( \sin n t \cdot \cos \frac{\pi}{3} + \cos n t \cdot \sin \frac{\pi}{3} \right) \] Since \( \cos \frac{\pi}{3} = \frac{1}{2} \) and \( \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2} \), substitute these values in: \[ x = 2 \left( \sin n t \cdot \frac{1}{2} + \cos n t \cdot \frac{\sqrt{3}}{2} \right) \] Simplify: \[ x = \sin n t + \sqrt{3} \cos n t \]

Given function: \[ y = 4 \sin \left(n t + \frac{\pi}{6}\right) \] Use the angle addition formula for sine: \[ \sin(A + B) = \sin A \cos B + \cos A \sin B \] Substitute \( A = n t \) and \( B = \frac{\pi}{6} \): \[ y = 4 \left( \sin n t \cdot \cos \frac{\pi}{6} + \cos n t \cdot \sin \frac{\pi}{6} \right) \] Since \( \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2} \) and \( \sin \frac{\pi}{6} = \frac{1}{2} \), substitute these values in: \[ y = 4 \left( \sin n t \cdot \frac{\sqrt{3}}{2} + \cos n t \cdot \frac{1}{2} \right) \] Simplify: \[ y = 2 \sqrt{3} \sin n t + 2 \cos n t \]

Use the equations found in Step 1 and Step 2: \[ x = \sin n t + \sqrt{3} \cos n t \] \[ y = 2 \sqrt{3} \sin n t + 2 \cos n t \] Isolate \( \sin n t \) and \( \cos n t \): From the equation for \(y\): \[ y = 2 \sqrt{3} \sin n t + 2 \cos n t \] Divide by 2: \[ \frac{y}{2} = \sqrt{3} \sin n t + \cos n t \] Subtract \( \cos n t \): \[ \frac{y}{2} - \cos n t = \sqrt{3} \sin n t \] Square both equations: \[ x^2 = (\sin n t + \sqrt{3} \cos n t)^2 \] \[ \left( \frac{y}{2} - \cos n t \right)^2 = (\sqrt{3} \sin n t)^2 \] Expand: \[ x^2 = \sin^2 n t + 2 \sin n t \cos n t \sqrt{3} + 3 \cos^2 n t \] \[ \left( \frac{y}{2} - \cos n t \right)^2 = 3 \sin^2 n t \] Substitute \( \cos^2 n t = 1 - \sin^2 n t \) and isolate variables: Combine and substitute to form the Cartesian equation.

Rewrite \( \cos^2 \theta \) in terms of \( \sin \theta \): \[ \cos^2 \theta = 1 - \sin^2 \theta \] Substitute into the given equation: \[ 3(1 - \sin^2 \theta) + 5 \sin \theta - 1 = 0 \] Simplify: \[ 3 - 3 \sin^2 \theta + 5 \sin \theta - 1 = 0 \] \[ -3 \sin^2 \theta + 5 \sin \theta + 2 = 0 \] Multiply by -1 to make it easier to solve: \[ 3 \sin^2 \theta - 5 \sin \theta - 2 = 0 \] Let \( u = \sin \theta \), then solve the quadratic equation: \[ 3u^2 - 5u - 2 = 0 \] Use the quadratic formula: \[ u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Where \( a = 3 \), \( b = -5 \), \( c = -2 \): \[ u = \frac{5 \pm \sqrt{25 + 24}}{6} \] \[ u = \frac{5 \pm 7}{6} \] Solve for the solutions: \[ u = 2 \] (not valid since \( \sin \theta \) can't be greater than 1), and \( u = -\frac{1}{3} \). So, \( \sin \theta = -\frac{1}{3} \). Find all solutions for \( \theta \) in the interval \( 0^\circ < \theta < 360^\circ \). \( \theta = 180 - \arcsin\left( -\frac{1}{3}\right) \) and \( \theta = 360 + \arcsin\left(- \frac{1}{3}\right) \).