5.4 Determine the general solution of: \( \cos 2 x-\frac{1}{3}=\frac{1}{3} \sin x \)

To find the general solution of the equation \( \cos 2x - \frac{1}{3} = \frac{1}{3} \sin x \), start by rewriting \( \cos 2x \) using the double-angle identity: \[ \cos 2x = 1 - 2\sin^2 x \] Substituting this into the equation gives: \[ 1 - 2\sin^2 x - \frac{1}{3} = \frac{1}{3} \sin x \] Now, simplify and rearrange the equation: \[ \frac{2}{3} - 2\sin^2 x = \frac{1}{3} \sin x \] Multiplying through by 3 to eliminate fractions results in: \[ 2 - 6\sin^2 x = \sin x \] Rearranging leads to a quadratic form: \[ 6\sin^2 x + \sin x - 2 = 0 \] Using the quadratic formula \( \sin x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 6, b = 1, c = -2 \): \[ \sin x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 6 \cdot (-2)}}{2 \cdot 6} = \frac{-1 \pm \sqrt{1 + 48}}{12} = \frac{-1 \pm 7}{12} \] This gives two potential solutions for \( \sin x \): 1. \( \sin x = \frac{6}{12} = \frac{1}{2} \) 2. \( \sin x = \frac{-8}{12} = -\frac{2}{3} \) Now, solve these sine values: 1. For \( \sin x = \frac{1}{2} \), the solutions are: \[ x = \frac{\pi}{6} + 2k\pi \quad \text{and} \quad x = \frac{5\pi}{6} + 2k\pi, \quad k \in \mathbb{Z} \] 2. For \( \sin x = -\frac{2}{3} \), use the inverse sine function: \[ x = \arcsin\left(-\frac{2}{3}\right) + 2k\pi \quad \text{and} \quad x = \pi - \arcsin\left(-\frac{2}{3}\right) + 2k\pi, \quad k \in \mathbb{Z} \] Thus, the general solutions can be gathered as: \[ x = \frac{\pi}{6} + 2k\pi, \quad x = \frac{5\pi}{6} + 2k\pi, \quad x = \arcsin\left(-\frac{2}{3}\right) + 2k\pi, \quad x = \pi - \arcsin\left(-\frac{2}{3}\right) + 2k\pi, \quad k \in \mathbb{Z} \]