5.4 Given: \( 1-\cos \theta=2 \sin ^{2} \theta \) 5.4.1 Show that the equation can be written as: \( (2 \cos \theta+1)(\cos \theta-1)=0 \). 5.4.2 Hence, determine the general solution of \( (2 \cos \theta+1)(\cos \theta-1)=0 \).

To show that \( 1 - \cos \theta = 2 \sin^2 \theta \), we start by using the identity \( \sin^2 \theta = 1 - \cos^2 \theta \). First, rewrite the equation as \( 1 - \cos \theta = 2(1 - \cos^2 \theta) \), then simplifying gives \( 1 - \cos \theta = 2 - 2 \cos^2 \theta \). Rearranging yields \( 2 \cos^2 \theta - \cos \theta - 1 = 0 \). This factors to \( (2 \cos \theta + 1)(\cos \theta - 1) = 0 \). Now, to find the general solution, we solve each factor separately. From \( 2 \cos \theta + 1 = 0 \), we have \( \cos \theta = -\frac{1}{2} \), which gives solutions \( \theta = \frac{2\pi}{3} + 2k\pi \) and \( \theta = \frac{4\pi}{3} + 2k\pi \), where \( k \) is any integer. For \( \cos \theta - 1 = 0 \), we get \( \cos \theta = 1 \), leading to the solution \( \theta = 2k\pi \). Hence, the general solutions are \( \theta = \frac{2\pi}{3} + 2k\pi, \, \theta = \frac{4\pi}{3} + 2k\pi, \, \text{and} \, \theta = 2k\pi \).