Determine the general solution of the following equation: \( 6 \sin ^{2} x+7 \cos x-3=0 \)

To solve the equation \( 6 \sin^{2} x + 7 \cos x - 3 = 0 \), we can use the Pythagorean identity \( \sin^{2} x = 1 - \cos^{2} x \). Substituting for \( \sin^{2} x \) gives us: \[ 6(1 - \cos^{2} x) + 7 \cos x - 3 = 0 \] This simplifies to: \[ 6 - 6 \cos^{2} x + 7 \cos x - 3 = 0 \] Which can be rewritten as: \[ -6 \cos^{2} x + 7 \cos x + 3 = 0 \] Multiplying the entire equation by -1 to make leading coefficient positive: \[ 6 \cos^{2} x - 7 \cos x - 3 = 0 \] Letting \( u = \cos x \), we have: \[ 6u^{2} - 7u - 3 = 0 \] Now, we can use the quadratic formula \( u = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} \): \[ u = \frac{-(-7) \pm \sqrt{(-7)^{2} - 4 \cdot 6 \cdot (-3)}}{2 \cdot 6} \] \[ u = \frac{7 \pm \sqrt{49 + 72}}{12} \] \[ u = \frac{7 \pm \sqrt{121}}{12} \] \[ u = \frac{7 \pm 11}{12} \] This gives us two solutions for \( u \): 1. \( u = \frac{18}{12} = \frac{3}{2} \) (not valid since \( \cos x \) cannot exceed 1) 2. \( u = \frac{-4}{12} = -\frac{1}{3} \) Now, we solve \( \cos x = -\frac{1}{3} \). The general solutions for \( x \) when \( \cos x = -\frac{1}{3} \) are given by: \[ x = \cos^{-1}(-\frac{1}{3}) + 2k\pi \quad \text{and} \quad x = -\cos^{-1}(-\frac{1}{3}) + 2k\pi \quad \text{for } k \in \mathbb{Z} \] Thus, the general solution for the given equation is: \[ x = \cos^{-1}(-\frac{1}{3}) + 2k\pi \quad \text{or} \quad x = -\cos^{-1}(-\frac{1}{3}) + 2k\pi, \quad k \in \mathbb{Z} \]