(ii) \( 5 \sin ^{2} y+9 \cos y-3=0 \) for \( 0^{\circ}

To solve the equation \( 5 \sin^2 y + 9 \cos y - 3 = 0 \), we can use the Pythagorean identity \( \sin^2 y = 1 - \cos^2 y \). This allows us to express everything in terms of \( \cos y \). 1. Substitute \( \sin^2 y \) with \( 1 - \cos^2 y \): \[ 5(1 - \cos^2 y) + 9 \cos y - 3 = 0 \] 2. Simplify the equation: \[ 5 - 5 \cos^2 y + 9 \cos y - 3 = 0 \] \[ -5 \cos^2 y + 9 \cos y + 2 = 0 \] 3. Multiply the entire equation by -1 to make the leading coefficient positive: \[ 5 \cos^2 y - 9 \cos y - 2 = 0 \] 4. Now, we can use the quadratic formula to solve for \( \cos y \). The quadratic formula is given by: \[ \cos y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] In our case, \( a = 5 \), \( b = -9 \), and \( c = -2 \). 5. Calculate the discriminant: \[ b^2 - 4ac = (-9)^2 - 4(5)(-2) = 81 + 40 = 121 \] 6. Now apply the quadratic formula: \[ \cos y = \frac{9 \pm \sqrt{121}}{2 \cdot 5} = \frac{9 \pm 11}{10} \] This gives us two possible solutions for \( \cos y \): \[ \cos y = \frac{20}{10} = 2 \quad \text{(not valid, since } \cos y \text{ must be in } [-1, 1]) \] \[ \cos y = \frac{-2}{10} = -0.2 \] 7. Now we need to find \( y \) such that \( \cos y = -0.2 \). The cosine function is negative in the second and third quadrants. 8. To find the reference angle, we calculate: \[ y_{\text{ref}} = \cos^{-1}(-0.2) \] Using a calculator: \[ y_{\text{ref}} \approx 102.0^\circ \] 9. Now, we find the angles in the second and third quadrants: - In the second quadrant: \[ y_1 = 180^\circ - y_{\text{ref}} \approx 180^\circ - 102.0^\circ \approx 78.0^\circ \] - In the third quadrant: \[ y_2 = 180^\circ + y_{\text{ref}} \approx 180^\circ + 102.0^\circ \approx 282.0^\circ \] 10. Therefore, the solutions for \( y \) in the interval \( 0^\circ < y < 360^\circ \) are: \[ y \approx 102.0^\circ \quad \text{and} \quad y \approx 282.0^\circ \] Thus, the final solutions are: \[ \boxed{102.0^\circ \text{ and } 282.0^\circ} \]